NCERT Solutions for Class 11 chemistry chapter 2 – Structure of Atom

Q.1  (i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons.

Ans.

(i) Mass of one electron = 9.10939 × 10^–31kg

Number of electrons that weigh 9.10939 × 10^–31kg = 1

Number of electrons that will weigh 1 g or 1 × 10^–3 kg

=\frac{1}{9.10939\times {10}^{-31}kg}\times 1\times {10}^{-3}kg

= 0.1098 × 10²⁸

= 1.098 x 10²⁷

(ii) Mass of one electron = 9.10939 × 10^–31kg

Mass of one mole of electron =

(6.022 × 10²³) ×(9.10939 ×10^–31kg)

= 5.48 × 10^–7kg

Charge on one electron = 1.6022 × 10^–19coulomb

Charge on one mole of electron

= (1.6022 × 10^–19C) (6.022 × 10²³)

= 9.65 × 10⁴C

Q.2 (i) Calculate the total number of electrons present in one mole of methane.

(ii) Find

(a) the total number and

(b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron =

1.675 × 10^-27kg).

(iii) Find

(a) the total number and

(b) the total mass of protons in 34 mg of NH₃ at STP. Will the answer change if the temperature and pressure are changed?

Ans.

(i) Number of electrons present in 1 molecule of methane (CH₄) = {1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.023 ×10²³ molecules of methane = 6.022 × 10²³× 10

= 6.022 × 10²⁴

(ii) (a) Number of atoms of ¹⁴C in 1 mole= 6.023 × 10²³

Since, 1 atom of ¹⁴C contains (14 – 6) i.e., 8 neutrons, the total number of neutrons in 14 g (or 1 mole) of ¹⁴C is (6.023 × 10²³) × 8.

Therefore, 14 g of ¹⁴C contains (6.023 × 10²³× 8) neutrons.

Therefore, number of neutrons in 7 mg of ¹⁴C is

=\frac{6.023\times {10}^{23}\times 8\times 7mg}{14000\text{mg}}

= 2.4092 × 10²¹

(b) Mass of one neutron = 1.67493 × 10^–27kg

Mass of total neutrons in 7 mg of ¹⁴C =

= (2.4092 × 10²¹) (1.67493× 10^–27kg)

= 4.0352 × 10^–6kg

(iii) (a) 1 mole of N of NH₃ = {1(14) + 3(1)} g of NH₃

= 17 g of NH₃ = 6.022× 10²³ molecules of NH₃

Total number of protons present in 1 molecule of NH₃

= {1(7) + 3(1)} =10

Number of protons in 6.023 × 10²³ molecules of NH₃

= (6.023 × 10²³) (10)

= 6.023 × 10²⁴

=17 g of NH₃ contains (6.023 × 10²⁴) protons

Number of protons in 34 mg of NH₃

=\frac{6.022\times {10}^{24}\times 34mg}{17000\text{mg}}

= 1.2046 × 10²²

(b) Mass of one proton = 1.67493 × 10^–27kg

Total mass of protons in 34 mg of NH₃

= (1.67493 × 10^–27kg) (1.2046 × 10²²)

= 2.0176 × 10^–5kg

The number of sub-atomic particles (protons, electrons, and neutrons) in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Q.3 How many neutrons and protons are there in the following nuclei?

{}_6^{13}\mathrm{C,}{}_8^{16}\mathrm{O,}{}_{12}^{24}M\mathrm{g,}{}_{26}^{56}F\mathrm{e,}{}_{38}^{88}Sr

Ans.

Atomic mass of

{}_6^{13}C

= 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass – Atomic number)

= 13 – 6 = 7

Atomic mass of

{}_8^{16}O

=16

Atomic number = 8 and Number of protons = 8

Number of neutrons = (Atomic mass – Atomic number)

= 16 – 8 = 8

Atomic mass of

{}_{12}^{24}Mg

=24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass – Atomic number)

= 24 – 12 = 12

Atomic mass of

{}_{26}^{56}Fe

=56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass – Atomic number)

= 56 – 26 = 30

Atomic mass of

{}_{38}^{88}Sr

=88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass – Atomic number)

= 88 – 38 = 50

Q.4 Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)

(i) Z = 17, A = 35

(ii) Z = 92, A = 233

(iii) Z = 4, A = 9

Ans.

Q.5 Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number

\left(\overline{v}\right)

of the yellow light.

Ans.

λ= c/ν

or, ν = c/ λ ——— (1)

Here,

ν = frequency of yellow light

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ= wavelength of yellow light = 580 nm = 580 × 10^–9m

Substituting the values in expression (1):

v=\frac{3\times {10}^8}{580\times {10}^{-9}}=5.17\times {10}^{14}{s}^{-1}

Thus, frequency of yellow light emitted from the sodium lamp is = 5.17 × 10¹⁴s^–1.

Wave number of yellow light, ν = 1/λ.

\overline{v}=\frac{1}{{\mathrm{580\; x\; 10}}^{-9}}=1.72\times {10}^6{m}^{-1}

Q.6 Find energy of each of the photons which

(i) correspond to light of frequency 3× 10¹⁵ Hz

(ii) have wavelength of 0.50 Å.

Ans.

(i) Energy (E) of a photon is given by the expression,

E = hν

Where,

h = Planck’s constant = 6.626 × 10^–34Js

ν= frequency of light = 3 × 10¹⁵Hz

Substituting the values in the given expression of E:

E= (6.626 x 10^-34) (3 x 10¹⁵)

=1.988 × 10^–18J

(ii) Energy (E) of a photon having wavelength (λ) is given by the expression,

E = h(c/λ)

h = Planck’s constant = 6.626 × 10^–34Js

c = velocity of light in vacuum = 3 × 10⁸m/s

λ =0.50 Å = 0.50 x 10^–10 m

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{0.50\times {10}^{-10}}=3.976\times {10}^{-15} E=3.98\times {10}^{-15}J

Q.7 Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10^–10 s.

Ans.

Frequency (ν) of light = 1/period

=\frac{1}{2.0\times {10}^{-10}s}=5.0\times {10}^9{s}^{-1}

Wavelength (λ) of light = c/ν

Where, c = velocity of light in vacuum = 3×10⁸ m/s

Substituting the value in the given expression of λ:

\lambda =\frac{3\times {10}^8}{5.0\times {10}^9}=6.0\times {10}^{-2}{m}^{\mathrm{}}

Wave number

\left(\overline{v}\right)

of light = 1/λ

=\frac{1}{6.0\times {10}^{-2}S}=1.66\times {10}^1{m}^{-1}=16.66m^{-1}

Q.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Ans.

Energy (E) of a photon = hν

Energy (E_n) of ‘n’ photons = nhν

n=E_n\lambda /hc

Where,

λ = wavelength of light = 4000 pm = 4000 ×10^–12m

c = velocity of light in vacuum = 3 × 10⁸ m/s

h = Planck’s constant = 6.626 × 10^–34 Js

Substituting the values in the given expression of n, i.e.,

n=E_n\lambda /hc

We get,

n=\frac{E_n\times \lambda }{h\times c}=\frac{1J\times 4\times {10}^{-9}m}{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}=2.012\times {10}^{16}photons

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 10¹⁶.

Q.9 A photon of wavelength 4 × 10^–7m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photo electron (1 eV= 1.6020 × 10^–19 J)

Ans.

(i) Energy (E) of a photon= hν= hc/λ

Where, h = Planck’s constant = 6.626 × 10^–34 Js

c = velocity of light in vacuum = 3 × 10⁸ m/s

λ = wavelength of photon = 4 × 10^–7 m

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{4\times {10}^{-7}}=4.9695\times {10}^{-19}J

Hence, the energy of the photon is 4.97 × 10^–19 J.

(ii) The kinetic energy of emission E_k is given by

= hν – hν₀ = (E–W) eV

Where W is the work function. Let us place value of E and W in electron volt in the above equation, we get

E_k=\left(\frac{4.9695\times {10}^{-19}}{1.6020\times {10}^{-19}}\right)eV-2.13eV

= (3.1020 – 2.13) eV

= 0.9720 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the following expression:

\begin{array}{l}\frac{1}{2}m{v}^2=h\nu -h{\nu }_0\\ or\text{v}=\sqrt{\left\{2\left(h\nu -h{\nu }_0\right)/m\right\}}\end{array}

Where, (hν – hν₀) is kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Multiply kinetic energy with 1.6020 × 10^–19 to convert into joules. Substitute these values in the given expression of v:

\begin{array}{l}v=\sqrt{\frac{2\times \left(0.9720\times 1.6020\times {10}^{-19}\right)J}{9.10939\times {10}^{-31}kg}}\\ =\sqrt{0.3418\times {10}^{12}{m}^2{S}^{-2}}\end{array}

v = 5.84 × 10⁵ ms^–1

Hence, the velocity of the photo electron is 5.84 × 10⁵ ms^-1.

Q.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^–1

Ans.

Energy of sodium (E) = N_Ahcλ

=\frac{\left({6.023510}^{23}mo{l}^{-1}\right)\left({6.626510}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{242\times {10}^{-9}m}

= 4.947 × 10⁵ J mol^–1

= 494 kJ mol^-1

Q.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 mm. Calculate the rate of emission of quanta per second.

Ans.

Power of bulb, P = 25 Watt = 25 Js^–1

Energy of one photon, E = hν = hc/λ

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{\left(0.57\times {10}^{-6}m\right)}

E = 34.87 × 10^–20 J

Rate of emission of quanta per second, R

=\frac{25}{34.87\times {10}^{-20}}=7.169\times {10}^{19}{s}^{-1}

Q.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν₀) and work function (W₀) of the metal.

Ans.

Threshold wavelength of radian (λ₀) =6800 Å

= 6800 × 10^–10 m

Threshold frequency (ν₀) of the metal= c/λ₀

=\frac{3\times {10}^{8}m{s}^{-1}}{6.8\times {10}^{-7}m}

= 4.41 × 10¹⁴ s^-1

Thus, the threshold frequency (ν₀) of the metal is 4.41 × 10¹⁴ s^–1

Hence, work function (W₀) of the metal = hν₀

=\left(6.626\times {10}^{-34}Js\right)\left(4.41\times {10}^{14}{s}^{-1}\right)=2.922\times {10}^{-19}J

Q.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Ans.

Transition of hydrogen atom from an energy level with n_i = 4 to n_f = 2 will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

\Delta E=2.18\times {10}^{-18}J\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)=h\nu

Substituting the values in the given expression of E:

\Delta E=2.18\times {10}^{-18}J\left(\frac{1}{4^2}-\frac{1}{2^2}\right) \Delta E=2.18\times {10}^{-18}J\left(\frac{\left(1-4\right)}{16}\right) \Delta E=2.18\times {10}^{-18}J\left(\frac{-3}{16}\right)

E = – (4.0875 × 10^–19J)

The negative sign indicates the energy of emission.

Wavelength of light emitted, λ = hc/E

Substitute these values in the given expression of λ:

\lambda =\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{\left(4.0875\times {10}^{-19}\right)} \lambda =486.3\times {10}^{-9}m=486nm

Q.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

Ans.

The expression of energy is given by,

E_n=\frac{-\left(2.18\times {10}^{-18}\right)Z^2}{n^2}

Z = atomic number of the atom

n = principal quantum number

For ionisation from n₁= 5 to n₂ = ∞

\Delta E=E_{\propto }-E_5

=\left(\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left(1^2\right)}{{(\propto )}^2}\right\}-\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left(1^2\right)}{5^2}\right\}\right)

=\left(2.18\times {10}^{-18}J\right)\left(\frac{1}{\left(5^2\right)}\right)Since,\frac{1}{\propto }=0

= 0.0872 x 10^-18 J

ΔE = 8.72 x 10^-20 J

Hence, the energy required for ionization from n = 5 to n = ∞ is 8.72 × 10^–20J.

Energy required for n₁ = 1 to n = ∞,

ΔE’ = E _∞ – E₁

=\left(\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left(1^2\right)}{{(\propto )}^2}\right\}-\left\{\frac{-\left(2.18\times {10}^{-18}J\right)\left(1^2\right)}{1^2}\right\}\right) =\left(2.18\times {10}^{-18}J\right)\left(1-0\right)

= 2.18 x 10^-18 J

On comparing the ionisation energies,

ΔE/ΔE’ =2.18 x 10^-18 J / 8.72 x 10^-20 J = 25

Thus, the energy required to remove an electron from n=1 orbit in a hydrogen atom is 25 times the energy required to remove an electron from n = 5 orbit.

Q.15 What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Ans.

When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible

Thus, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The numbers of spectral lines produced when an electron in the n^th level drops down to the ground state is given by n (n – 1)/2.

When n = 6, the Number of spectral lines = 6 (6 – 1)/2 = 15

Q.16

(i) The energy associated with the first orbit in the hydrogen atom is –2.18 ×10^–18 J atom^–1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Ans.

(i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

E_5=\frac{-\left(2.18\times {10}^{-18}\right)}{5^2}=\frac{-2.18\times {10}^{-18}}{25}

E₅ = – 8.72 × 10^–20 J

(ii) Radius of Bohr’s n^th orbit for hydrogen atom is given by,

r_n = (0.0529 nm) n²

For n = 5, let us calculate radius for hydrogen atom for fifth orbit

r₅ = (0.0529 nm) (5)²

r₅ = 1.3225 nm

Q.17 Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans.

For the Balmer series, n_i = 2. Thus, the expression of wave number

\left(\overline{v}\right)

is given by

\overline{v}=\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{n_f^2}\right)\left(1.097\times {10}^7\right)m^{-1} \overline{v}=1/\lambda

Wave number is inversely proportional to wavelength of transition. Hence, the longest wavelength transition, wave number has to be the smallest.

For wave number to be minimum, n_f should be minimum. For the Balmer series, a transition from n_i = 2 to n_f = 3 is allowed. Hence, by taking n_f = 3, we get:

\begin{array}{l}\overline{v}=\left(1.097\times {10}^7\right)\left(\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right)\\ \overline{v}=\left(1.097\times {10}^7\right)\left(\frac{1}{4}-\frac{1}{9}\right)\\ \overline{v}=\left(1.097\times {10}^7\right)\left(\frac{9-4}{36}\right)\end{array}

ν = 1.5236 × 10⁶ m^–1

Q.18 What is the energy in Joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10^–11 ergs.

Ans.

Let us convert ground state electron energy into joules.

Ground state electron energy = – 2.18 × 10^–11 ergs

= – 2.18 × 10^–11 × 10^–7 J

= – 2.18 × 10^–18 J

Energy (E) of the n^th Bohr orbit of an atom is given by,

E_n=\frac{-\left(2.18\times {10}^{-18}\right)Z^2}{n^2}

Where, Z = atomic number of the atom

Energy required for shifting the electron from n = 1 to n = 5 is given as:

ΔE = E₅ -E₁

=\frac{-\left(2.18\times {10}^{-18}\right){\left(1\right)}^2}{5^2}-\left(2.18\times {10}^{-18}\right) =\left(2.18\times {10}^{-18}\right)\left(1-\frac{1}{25}\right) =\left(2.18\times {10}^{-18}\right)\left(\frac{24}{25}\right)

= 2.0928 x 10^-18 J

Wave length of emitted light = hc/E

=\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{\left(2.0928\times {10}^{-18}\right)}

= 9.498 x 10^-8 m

Q.19 The electron energy in hydrogen atom is given by E_n = (–2.18 × 10^–18)/n² J. Calculate the energy required to remove an electron completely from the n= 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Ans.

Given,

E_n=-\frac{2.18\times {10}^{-18}}{n^2}J

Energy required for ionization from n = 2 is given by,

= 0.545 x 10^-18 J

\Delta E=E_{\propto }-E_2 =\left(\left\{\frac{-\left(2.18\times {10}^{-18}J\right)}{{(\propto )}^2}\right\}-\left\{\frac{-\left(2.18\times {10}^{-18}J\right)}{2^2}\right\}\right) =\left(\frac{2.18\times {10}^{-18}}{4}-0\right)

ΔE = 5.45 x 10^-19 J

λ = hc/ΔE

Here, λ is the longest wavelength causing the transition.

\lambda =\frac{\left(6.626\times {10}^{-34}\right)\left(3\times {10}^8\right)}{\left(5.45\times {10}^{-19}\right)}=3.647\times {10}^{-7}m

= 3647 x 10^-10 m

= 3647 Å

Q.20 Calculate the wavelength of an electron moving with a velocity of 2.05 ×10⁷ ms^–1.

Ans.

According to de Broglie’s equation,

λ = h/mv

Where,

λ = wavelength of moving particle

m = mass of particle

v= velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

\begin{array}{l}\lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(2.05\times {10}^{7}m{s}^{-1}\right)}\\ \lambda =3.548\times {10}^{-11}m\end{array}

Hence, the wavelength of the electron moving with a velocity of 2.05 × 10⁷ ms^–1 is 3.548 × 10^–11 m.

Q.21 The mass of an electron is 9.1 × 10^–31 kg. If its K.E. is 3.0 × 10^–25 J, calculate its wavelength.

Ans.

From de Broglie’s equation,

λ = h/mv

Given,

Kinetic energy (K.E.) of the electron = 3.0 × 10^–25 J.

K.E = ½ mv²

\begin{array}{l}Velocity(v)=\sqrt{2K.E./m}\\ =\sqrt{\frac{2\left(3.0\times {10}^{25}J\right)}{9.10939\times {10}^{-31}kg}}\end{array}

v = 812 ms^-1

Substituting the value in the expression of λ:

\lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(812m{s}^{-1}\right)}

λ= 8.9627 x 10^-7 m

Hence, the wavelength of the electron is 8.9627 × 10^–7 m.

Q.22 Which of the following are isoelectronic species, i.e., those having the same number of electrons? Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2–, Ar

Ans.

Number of electrons in Na⁺ = 11–1 = 10

K⁺ = 19–1 = 18

Mg²⁺= 12–2 = 10

Ca²⁺ = 20–2 = 18

S^2– = 16+2 = 18

Ar = 18

Hence Na⁺ and Mg²⁺ are isoelectronic species and

K⁺, Ca²⁺, S^2–, Ar are isoelectronic species.

Q.23 (i)Write the electronic configurations of the following ions: (a) H^– (b) Na⁺(c) O^2–(d) F^–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s¹ (b) 2p³ and (c) 3p⁵?

(iii)Which atoms are indicated by the following configurations?

(a) [He] 2s¹

(b) [Ne] 3s²3p³

(c) [Ar] 4s²3d¹.

Ans.

(i) (a) H^– ion:

The electronic configuration of H atom is 1s¹.

A negative charge on hydrogen indicates the gain of electron by it.

Electronic configuration of H^– = 1s²

(b) Na⁺ ion:

The electronic configuration of Na atom is 1s²2s²2p⁶3s¹.

A positive charge on the species indicates the loss of an electron by it.

Electronic configuration of Na⁺ =1s² 2s² 2p⁶ 3s⁰

=1s² 2s² 2p⁶

(c)O^2– ion:

The electronic configuration of O atom is 1s² 2s² 2p⁴

A dinegative charge on the species indicates that two electrons are gained by it.

Electronic configuration of O^2–ion = 1s² 2s² 2p⁶

(d)F^–ion:

The electronic configuration of F atom is 1s² 2s² 2p⁵.

A negative charge on the species indicates the gain of an electron by it.

Electron configuration of F^– ion = 1s² 2s² 2p⁶

(ii) (a) 3s¹:

Completing the electron configuration of the element as = 1s² 2s² 2p⁶ 3s¹

Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11

Atomic number of the element = 11

(b)2p³:

Completing the electron configuration of the element as

= 1s²2s²2p³

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c) 3p⁵:

Completing the electron configuration of the element as

= 1s²2s²2p⁵

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(iii) (a)[He] 2s¹:

The electronic configuration of the element is

[He] 2s¹= 1s²2s¹.

Atomic number of the element = 3

The element is lithium (Li).

b)[Ne] 3s²3p³ :

The electronic configuration of the element is

[Ne]3s²3p³ = 1s²2s²2p⁶3s²3p³

Atomic number of the element = 15

Hence, the element with the electronic configuration

[Ne] 3s²3p³ is phosphorus (P).

(c)[Ar] 4s²3d¹:

The electronic configuration of the element is [Ar]4s²3d¹= 1s²2s²2p⁶3s²3p⁶4s²3d¹.

Atomic number of the element = 21

Hence, the element with the electronic configuration

[Ar] 4s²3d¹ is scandium (Sc).

Q.24 What is the lowest value of n that allows g- orbitals to exist?

Ans.

For g-orbital, l (Azimuthal quantum number) = 4,

As for any value of principal quantum number(n), the Azimuthal quantum number (l) can have a value from zero to (n–1).

∴ For l= 4, minimum value of n= 5

Q.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and m_l for this electron.

Ans.

For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic quantum number (m_l) = –2,–1, 0, 1, 2

Q.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce

(i) the number of protons and (ii) the electronic configuration of the element.

Ans.

(i) For a neutral atom, number of protons is equal to the number of electrons. Thus the number of protons in the atom of given element = 29.

(ii)The electronic configuration of the atom is

= 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰.

Q.27 Give the number of electrons in the species H₂⁺, H₂, and O₂⁺.

Ans.

Number of electrons present in hydrogen molecule (H₂)

=1+1 = 2

Number of electrons in H₂⁺ = 2-1= 1

Number of electrons in H₂= 1 + 1 = 2

Number of electrons present in oxygen molecule (O₂) = 8 + 8 = 16

Number of electrons in O₂⁺ = 16 – 1 = 15

Q.28 (i) An atomic orbital has n = 3. What are the possible values of l and m_l?

(ii) List the quantum numbers (m_l and l) of electrons for 3d orbital.

(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

Ans.

(i) n = 3 (Given)

For a given value of n, l can have values from 0 to

(n– 1).

For n= 3, l= 0, 1, 2

For a given value of l, m_l can have (2l+ 1) values

For l= 0, m= 0

l= 1, m= – 1, 0, 1

l= 2, m= – 2, – 1, 0, 1, 2

Thus, for n= 3 l= 0, 1, 2 m₀= 0

m₁= – 1, 0, 1 m₂= – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l= 2. For a given value of l, m_l can have (2l+ 1) values i.e., 5 values.

∴ For l = 2, m_l = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l= 1.

For a given value of n, l can have values from zero to (n– 1).

∴ For l =1, the minimum value of n is 2.

Similarly, for f-orbital, l = 3.

For l= 3, the minimum value of n is 4.

Hence, 1p and 3f do not exist.

Q.29 Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n= 1, l = 0; (b)n= 3; l=1 (c)n = 4; l= 2; (d)n= 4; l=3

Ans.

(a) For n= 1, l= 0 (Given), the orbital is 1s.

(b)For n= 3 and l= 1, the orbital is 3p.

(c) For n= 4 and l= 2, the orbital is 4d.

(d) For n= 4 and l= 3, the orbital is 4f.

Q.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.

\begin{array}{l}a)n=0l=0m_l=0m_s=+1/2\\ b)n=1l=0m_l=0m_s=-1/2\\ c)n=1l=1m_l=0m_s=+1/2\\ d)n=2l=1m_l=0m_s=-1/2\\ e)n=3l=3m_l=-3m_s=+1/2\\ f)n=3l=1m_l=0m_s=+1/2\end{array}

Ans.

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c)The given set of quantum numbers is not possible.

For a given value of n, ‘l’ can have values from zero to (n– 1). For n= 1, l= 0 and not 1.

(d)The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible.

For n= 3, l= 0 to (3 – 1),i.e., l= 0 to 2 or 0, 1, 2

(f)The given set of quantum numbers is possible.

Q.31 How many electrons in an atom may have the following quantum numbers?

(a) n= 4, m_s=–1/2 and (b) n=3, l=0

Ans.

(a) Total number of electrons in an atom for a value of n= 2n²

For n=4, total number of electrons= 2(4)² =32.

The given element has a fully filled orbital as

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰

Hence, all the electrons are paired.

Number of electrons (having n= 4 and m_s=–1/2) = 16

(b) n = 3, l= 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n= 3 and l= 0 is 2.

Q.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Ans.

Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of electron is given by:

mvr=nh/2\pi \mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots }(1)

Where,

n = 1, 2, 3…

According to de Broglie’s equation:

\begin{array}{l}\lambda =h/mv\\ Or,\text{mv}=h/\lambda \mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots }(2)\end{array}

Substituting the value of ‘mv’ from equation (2) in equation (1):

\begin{array}{l}hr/\lambda =nh/2\pi \\ Or,\text{2}\pi \text{r}=n\lambda \mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots }(3)\end{array}

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

Q.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n= 4 to n= 2 of He⁺ spectrum?

Ans.

For He⁺ ion, expression of the wave number

\left(\overline{v}\right)

associated with the Balmer transition, n=4 to n=2

Where, n₁= 2, and n₂=4

Z = atomic number of helium

\overline{v}=\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) =4R\frac{\left(4-1\right)}{16} \begin{array}{l}\overline{v}=1/\lambda =3R/4\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots }(2)\\ \lambda =4/3R\end{array}

Comparing above two equations the desired transition for hydrogen will have the same wavelength as that of He⁺.

\begin{array}{l}=R{\left(1\right)}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3R}{4}\\ =\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{3}{4}\mathrm{\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots }(1)\end{array}

By hit and trail method, the equality given by equation (1) is true only when

n₁= 1 and n₂= 2.

The transition for n₂= 2 to n₁= 1 in hydrogen spectrum would have the same wavelength as Balmer transition n= 4 to n = 2 of He⁺ spectrum.

Q.34 Calculate the energy required for the process

H{e}^{+}(g)\to H{e}^{2+}(g)+e^{-}

The ionization energy for the H atom in the ground state is 2.18 × 10^–18 J atom^–1.

Ans.

Energy associated with hydrogen-like species is given by,

E_n=-2.18\times {10}^{-18}\left(Z^2/n^2\right)J

For ground state of hydrogen atom,

ΔE = E _∞–E₁

=0-\left[-2.18\times {10}^{-18}\left\{\frac{{\left(2\right)}^2}{{\left(1\right)}^2}\right\}\right]J

ΔE= 2.18x 10^-18 J

For the given process,

H{e}^{+}(g)\to H{e}^{2+}(g)+e^{-}

An electron is removed from n= 1 to n= ∞.

ΔE = E _∞-E₁

=0-\left[-2.18\times {10}^{-18}\left\{\frac{{\left(2\right)}^2}{{\left(1\right)}^2}\right\}\right]J

ΔE= 8.72 x 10^-18 J

The energy required for the process= 8.72 x 10^-18 J.

Q.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Ans.

1 m= 100 cm, or 1cm= 10^-2 m

Length of the scale = 20 cm= 20 × 10^–2m

Diameter of a carbon atom = 0.15 nm= 0.15 × 10^–9m

One carbon atom occupies 0.15 × 10^–9m

Number of carbon atoms that can be placed in a straight line

=\frac{20\times {10}^{-2}m}{0.15\times {10}^{-9}m}=133.33\times {10}^7

= 1.33 x 10⁹

Q.36 2 × 10⁸atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Ans.

Number of carbon atoms present = 2 × 10⁸
Length of the given arrangement = 2.4 cm
Diameter of carbon atom

=\frac{2.4\times {10}^{-2}m}{2\times {10}^{8}m}=1.2\times {10}^{-10}

Radius of carbon atom = (Diameter/2)

=\frac{1.2\times {10}^{-10}m}{2}=6.0\times {10}^{-11}m

Q.37 The diameter of zinc atom is 2.6Å Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

Ans.

(a) Radius of zinc = (Diameter/2)

=2.6/2 Å

\begin{array}{l}=1.3\times {10}^{-10}m\\ =130\times {10}^{-12}m=130pm\end{array}

(b) Length of the arrangement = 1.6 cm

= 1.6 x 10^-2 m Diameter of zinc atom = 2.6 × 10^–10m

Number of zinc atoms present in the arrangement

\begin{array}{l}=\frac{1.6\times {10}^{-2}m}{2.6\times {10}^{-10}m}\\ =0.6153\times {10}^{8}m\\ =6.153\times {10}^{7}m\end{array}

Q.38 A certain particle carries 2.5 × 10^–16C of static electric charge. Calculate the number of electrons present in it.

Ans.

Charge on one electron = 1.6022 × 10^–19 C

= 1.6022 ×10^–19 C charge is carried by 1 electron

Number of electrons carrying a charge of 2.5 × 10^–16 C

=\frac{2.5\times {10}^{-16}C}{1.6022\times {10}^{-19}C}=1.560\times {10}^{3}C=1560C

Q.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10^–18C, calculate the number of electrons present on it.

Ans.

Charge on the oil drop = – 1.282 ×10^–18 C

Charge on one electron = – 1.6022 × 10^–19 C

∴Number of electrons present on the oil drop

\begin{array}{l}=\frac{-1.282\times {10}^{-18}C}{-1.6022\times {10}^{-19}C}\\ =0.8001\times {10}^1=8.0\end{array}

Q.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Ans.

A thin foil of lighter atoms cannot give the same results as given with the foil of heavier atoms.

The lighter atoms will not cause enough deflection of α- particles because they would be able to carry very little positive charge.

Q.41 Symbols

{}_{35}^{79}Br\text{and}{}^{79}Br

can be written, whereas symbols

{}_{79}^{35}B\mathrm{r\; and}{}^{35}Br

are not acceptable. Answer briefly.

Ans.

The convenient way of representing an element with its atomic mass (A) and atomic number (Z) is

{}_Z^{A}X

Atomic mass of Br is 79, but atomic number Z is 35, thus the symbol

{}_{35}^{79}Br

is acceptable but

{}_{79}^{35}Br

is not acceptable.

Again, the atomic number of an element is constant, but the atomic mass of an element depends

upon the relative abundance of its isotopes, so

{}^{79}Br

can be written but

{}^{35}Br

can not be written.

Q.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans.

Let the number of protons in the element be x. Number of neutrons in the element = x + 31.7% of x
= x + 0.317 x = 1.317 x
According to the question,
Mass number of the element = 81
(Number of protons + number of neutrons) = 81

X=\frac{81}{2.317}=34.95

X= 35
The number of protons in the element is 35.
Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.
The atomic symbol of the element is

{}_{35}^{81}Br

Q.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Ans.

Let the number of electrons in the ion carrying a negative charge be x.

Then, number of neutrons present = x+ 11.1% of x

= x + 0.111x = 1.111x

Number of electrons in the neutral atom = (x– 1)

(When an ion carries a negative charge, it carries an extra electron).

Number of protons in the neutral atom = (x – 1)

Mass number of the ion = 37

So, (x– 1) + 1.111x= 37

Or, 2.111x= 38 or, x= 18

The symbol of the ion is

{}_{17}^{37}C{l}^{-}

Q.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Ans.

Let the number of electrons present A³⁺ be x.
Number of neutrons in it = x + 30.4% of x = 1.304x
Since the ion is tri-positive,
Number of electrons in neutral atom = x+ 3 = Number of protons in neutral atom
Mass number of the ion = 56

\begin{array}{l}\left(x+3\right)+\left(1.304x\right)=56\\ or,2.304x=53\\ X=\frac{53}{2.304}or,X=23\end{array}

Number of protons = x + 3 = 23 + 3 = 26
The symbol of the

{}_{26}^{56}F{e}^{3+}

Q.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Ans.

The increasing order of frequency is as follows:

Radiation from FM radio < radiation from microwave oven < amber light < X- rays < Cosmic rays

Frequency is inversely proportional to the wavelength.

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < amber light < radiation from microwave ovens < radiation of FM radio.

Q.46 Nitrogen laser produces a radiation at a wave length of 337.1 nm. If the number of photons emitted is 5.6 × 10²⁴, calculate the power of this laser.

Ans.

Power of laser = Energy with which it emits photons

=E=\frac{Nhc}{\lambda }

Where,

N= number of photons emitted

h = Planck’s constant

c = velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression of Energy (E):

E=\frac{\left(5.4\times {10}^{24}\right)\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{337.1\times {10}^{-9}m}

Energy (E):
E = 0.32 × 10⁷J
= 3.2 × 10⁶J
The power of the laser is 3.2 × 10⁶J.

Q.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Ans.

Wavelength of radiation emitted = 616 nm

= 616 × 10^–9m (Given)

\nu =c/\lambda

Where,

c= velocity of radiation

λ = wavelength of radiation

Substituting the values in the given expression (ν)

\begin{array}{l}\nu =\frac{3.0\times {10}^{8}m/s}{616\times {10}^{-9}m}\\ =4.87\times {10}^8\times {10}^9\times {10}^{-3}{s}^{-1}\end{array}

= 4.87 × 10¹⁴s^–1

Frequency of emission (ν) = 4.87 × 10¹⁴s^-1

(b)\text{Velocity of radiation},\left(\text{c}\right)=\text{3}.0\times \text{1}{0}^{\text{8}}{\text{ms}}^{–\text{1}}

Distance travelled by this radiation in 30 s
= (3.0 × 10⁸ms^–1) (30 s)
= 9.0 × 10⁹m
(6.626 × 10^–34Js) (4.87 × 10¹⁴s^–1)
(c) Energy of quantum (E) = 32.27 × 10^–20J

=

\begin{array}{l}\\ (d)\text{Energy of one photon}\left(\text{quantum}\right)=\text{32}.\text{27}\times \text{1}{0}^{–\text{2}0}\text{J}\end{array}

Therefore, 32.27 × 10^–20J of energy is present in 1 quantum.
Number of quanta in 2 J of energy =

\nu =\frac{2J}{32.27\times {10}^{-20}J}

= 6.19 x 10¹⁸ = 6.2 x 10¹⁸

Q.48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10^–18J from the radiations of 600 nm, calculate the number of photons received by the detector.

Ans.

From the expression of energy of one photon (E)

E=hc/\lambda

Where,

λ = wavelength of radiation

h= Planck’s constant

c= velocity of radiation

Substituting the values in the given expression of E:

E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{600\times {10}^{-9}m}

E = 3.313 x 10^-19 J
Energy of one photon = 3.313 × 10^–19J
Number of photons received with 3.15 × 10^–18J energy

=\frac{315\times {10}^{-18}J}{3.313\times {10}^{-19}J}

=9.5

Q.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 10¹⁵, calculate the energy of the source.

Ans.

Frequency of radiation (ν) = 1/ (2.0x 10^-9 s)

= 5.0 x 10⁸ s^-1

Energy (E) of source = N h ν

Where,

N = number of photons emitted

h = Planck’s constant

ν = frequency of radiation

Substituting the values in the given expression of (E):

E= (2.5 × 10¹⁵) (6.626 × 10^–34Js) (5.0 × 10⁸s^–1)

= 8.282 × 10^–10J

Thus the energy of the source = 8.282x 10^-10 J

Q.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Ans.

λ₁= 589nm= 589 x 10^-9 m

λ₂ =589.6nm= 589.6 x 10^-9 m

\begin{array}{l}{v}_1=c/{\lambda }_1=\frac{3.0\times {10}^{8}m{s}^{-1}}{589\times {10}^{-9}m}=5.093\times {10}^{14}{s}^{-1}\\ v_2=c/{\lambda }_2=\frac{3.0\times {10}^{8}m{s}^{-1}}{589\times {10}^{-9}m}=5.088\times {10}^{14}{s}^{-1}\end{array}

ΔE= E₂ – E₁
= h (ν₂ – ν₁)
= (6.626 x 10^-34 Js) (5.093 – 5.088) x 10¹⁴ s^-1
= 3.31 x 10^-22 J

Q.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. (c) If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Ans.

Work function (W₀) for caesium atom is 1.9 eV.

(a)From the expression

W₀ = hc/λ₀

we get,

{\lambda }_0=\text{hc}/{\text{W}}_0

λ₀= threshold wavelength

h = Planck’s constant

c = velocity of radiation

Substituting the values in the given expression of (λ₀):

We get,

\begin{array}{l}{\lambda }_0=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{1.9\times 1.602\times {10}^{-19}J}\\ {\lambda }_0=6.53\times {10}^{-7}m\end{array}

The threshold wavelength is 653 nm.

(b) From the expression W₀ = hν₀, we get

ν₀ = W₀/h

Where, ν₀ = Threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν₀:

v_0=\frac{1.9\times 1.602\times {10}^{-19}J}{6.626\times {10}^{-34}Js}

(1 eV = 1.602 × 10^–19J)

ν₀ = 4.593 × 10¹⁴ s^–1

The threshold frequency of radiation (ν₀) is

4.593 × 10¹⁴ s^–1.

(c) According to the question:

Wavelength used in irradiation (λ) = 500 nm

Kinetic energy of ejected electron = h (ν–ν₀)

\begin{array}{l}=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)\\ =\left(6.626\times {10}^{-34}Js\right)\left(3.0\times {10}^{8}m{s}^{-1}\right)\left(\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right)\\ =\left(1.9878\times {10}^{-26}Jm\right)\left[\frac{\left(653-500\right)\times {10}^{-9}m}{\left(653\right)\left(500\right)\times {10}^{-18}{m}^2}\right]\\ =\frac{\left(1.9878\times {10}^{-26}Jm\right)\left(153\times {10}^9\right)}{\left(653\right)\left(500\right)}J\end{array}

= 9.3149 × 10^–20 J

Kinetic energy of the ejected photoelectron

= 9.3149 × 10^–20J

Since, K.E. = ½ mv²

= 9.3149 x 10^-20 J

\begin{array}{l}v=\sqrt{\frac{2\left(9.3149\times {10}^{-20}J\right)}{9.10939\times {10}^{-31}kg}}\\ =\sqrt{2.0451\times {10}^{11}{m}^2{s}^{-2}}\end{array}

v = 4.52 × 10⁵ ms^–1

The velocity of the ejected photoelectron is 4.52×10⁵ms^–1

Q.52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

λ (nm)	500	450	400
v x 10-5 (cm s-1)	2.55	4.35	5.35

Ans.

(a) Assuming the threshold wavelength

\begin{array}{l}{\lambda }_{0}nm=\left({\lambda }_0\times {10}^{-9}m\right),\\ h\left(\nu -{\nu }_0\right)=\frac{1}{2}m{v}^2\end{array}

Three different equalities can be formed by the given values as:

\begin{array}{l}=hc\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right)=\frac{1}{2}m{v}^2\\ hc\left(\frac{1}{500\times {10}^{-9}}-\frac{1}{{\lambda }_0\times {10}^{-9}}\right)=1/2m{\left(2.55\times {10}^{+3\mathrm{}}m{s}^{-1}\right)}^2\\ \frac{hc}{{10}^{-9}}\left(\frac{1}{500}-\frac{1}{{\lambda }_0}\right)=1/2m{\left(2.55\times {10}^{+3}m{s}^{-1}\right)}^2\mathrm{\dots \dots \dots \dots \dots \dots .}(1)\\ Similarly,\\ \frac{hc}{{10}^{-9}}\left(\frac{1}{450}-\frac{1}{{\lambda }_0}\right)=1/2m{\left(4.35\times {10}^{+3}m{s}^{-1}\right)}^2\mathrm{\dots \dots \dots \dots \dots \dots .}(2)\\ \frac{hc}{{10}^{-9}}\left(\frac{1}{400}-\frac{1}{{\lambda }_0}\right)=1/2m{\left(5.35\times {10}^{+3}m{s}^{-1}\right)}^2\mathrm{\dots \dots \dots \dots \dots \dots .}(3)\\ Dividng\text{equation (3) by equation (1):}\\ \frac{\left[\frac{{\lambda }_0-400}{400{\lambda }_0}\right]}{\left[\frac{{\lambda }_0-500}{500{\lambda }_0}\right]}=\frac{{\left(5.35\times {10}^{3}m{s}^{-1}\right)}^2}{{\left(2.55\times {10}^{3}m{s}^{-1}\right)}^2}\\ \frac{\left(5{\lambda }_0-2000\right)}{4{\lambda }_0-2000}={\left(\frac{5.35}{2.55}\right)}^2=\frac{28.6225}{6.5025}\\ \frac{\left(5{\lambda }_0-2000\right)}{4{\lambda }_0-2000}=4.40177\\ \end{array}

Solving this equation we get the value of λ₀ = 539.68 » 540 nm

(b) Substituting this value in equation (3),we get,

\begin{array}{l}\frac{h\times 3\times {10}^8}{{10}^{-9}}\left(\frac{1}{400}-\frac{1}{540}\right)=\frac{1}{2}\left(9.11\times {10}^{-31}\right){\left(5.35\times {10}^3\right)}^2\\ h=6.70\times {10}^{-38}Js\end{array}

Note: The value of h is not the correct value of Planck’s constant. It is solved according to the values given in NCERT. The correct value of Planck’s constant is

6.626\times {10}^{-34}Js

Q.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Ans.

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W₀) of radiation and its kinetic energy (K.E) i.e.,
E = W₀ + K.E.
Or, W₀ = E – K.E.
Energy of incident photon (E)= hc/λ
Where,
c = velocity of radiation
h = Planck’s constant
λ= wavelength of radiation
Substituting the values in the given expression of E:

\begin{array}{l}E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{256.7\times {10}^{-9}m}\\ =7.744\times {10}^{-19}J\\ =\frac{7.744\times {10}^{-19}}{1.602\times {10}^{-19}}eV\end{array}

E= 4.83 eV.
The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,
K.E = 0.35 V
Work function, W₀= E – K.E = 4.83 eV – 0.35 eV
= 4.48 eV

Q.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity of 1.5 × 10⁷ms^–1, calculate the energy with which it is bound to the nucleus.

Ans.

Energy of incident photon (E) is given by,

\begin{array}{l}E=hc/\lambda \\ E=\frac{\left(6.626\times {10}^{-34}Js\right)\left(3\times {10}^{8}m{s}^{-1}\right)}{150\times {10}^{-12}m}\end{array}

= 1.3252 x 10^-15 J = 13.252 x 10^-16 J
Energy of the electron ejected (K.E) = ½ mv²
= ½ (9.10939 x 10^-31 kg)(1.5 x 10⁷ ms^-1)²
= 10.2480 × 10^–17 J
= 1.025 × 10^–16 J
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E.
= 13.252 × 10^–16 J – 1.025 × 10^–16 J = 12.227 × 10^–16 J

=\frac{12.227\times {10}^{-16}}{1.602\times {10}^{-19}}eV

= 7.6 x 10³ eV

Q.55 Emission transitions in the Paschen series end at orbit n= 3 and start from orbit n and can be represented as ν= 3.29 × 10¹⁵(Hz) [1/3²– 1/n²] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Ans.

Q.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Ans.

The radius of the n^th orbit of hydrogen-like particles is given by,

\begin{array}{l}r=\frac{0.529n^2}{Z}\AA \\ r=\frac{52.9n^2}{Z}pm\end{array}

For radius (r₁) = 1.3225 nm = 1.3225 × 10^–9m
= 1322.5 × 10^–12m = 1322.5 pm

\begin{array}{l}{n}_1^2=\frac{r_{1}Z}{52.9}\\ Or,n_1^2=\frac{1322.5Z}{52.9}\\ Similarly\\ n_2^2=\frac{211.6Z}{52.9}\\ \frac{n_1^2}{n_2^2}=\frac{1322.5}{211.6}\\ \frac{n_1^2}{n_2^2}=6.25\\ \frac{n_1}{n_2}=2.5\\ \frac{n_1}{n_2}=\frac{25}{10}=\frac{5}{2}\end{array}

Or, n₁ = 5 and n₂ = 2
Thus, the transition is from the 5^th orbit to the 2^nd orbit. It belongs to the Balmer series.
Wave number

\left(\overline{v}\right)

or the transition is given by,

\begin{array}{l}=1.097\times {10}^7{m}^{-1}\left(\frac{1}{2^2}-\frac{1}{5^2}\right)\\ =1.097\times {10}^7{m}^{-1}\left(\frac{21}{100}\right)\end{array}

= 2.3037 x 10⁶ m^-1
Wavelength (λ) associated with the emission transition is given by,
λ =1
/

=\frac{1}{2.303\times {10}^6{m}^{-1}}

= 0.434 ×10^–6 m
λ = 434 nm
It lies in the visible region.

Q.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is 1.6 × 10⁶ms^–1, calculate de Broglie wavelength associated with this electron.

Ans.

From de Broglie’s equation,

\begin{array}{l}\lambda =h/mv\\ \lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(1.6\times {10}^{6}m{s}^{-1}\right)}\end{array}

λ= 4.55 x 10^-10 m, or λ = 455 pm
de Broglie’s wavelength associated with the electron is 455 pm.

Q.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Ans.

From de Broglie’s equation,

\begin{array}{l}\lambda =h/mv\\ v=h/m\lambda \end{array}

Where,
v= velocity of particle (neutron)
h= Planck’s constant
m= mass of particle (neutron)
λ = wavelength
Substituting the values in the expression of velocity (v),

v=\frac{6.626\times {10}^{-34}Js}{\left(1.67493\times {10}^{-27}kg\right)\left(800\times {10}^{-12}m\right)}

= 4.94 × 10²ms^–1
V = 494 ms^–1
Velocity associated with the neutron = 494 ms^–1

Q.59 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 10⁵ms^–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Ans.

Q.60 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is

h/4π_m×0.05 nm, is there any problem in defining this value?

Ans.

From Heisenberg’s uncertainty principle,

\Delta x\times \Delta p=h/4\pi \to \Delta P=\frac{1}{\Delta x}\times \frac{h}{4\pi }

Where,
Δx= uncertainty in position of the electron
Δp= uncertainty in momentum of the electron
Substituting the values in the expression in the last equation:

\begin{array}{l}\Delta p=\frac{1}{0.002nm}\times \frac{6.626\times {10}^{-34}Js}{4\times 3.14}\\ \Delta p=\frac{1}{0.002nm}\times \frac{6.626\times {10}^{-34}Js}{4\times 3.14}\end{array}

= 2.637 × 10^–23Jsm^–1
Δp = 2.637 × 10^–23 kgms^–1
Or, uncertainty in the momentum of the electron
= 2.637× 10^–23 kgms^–1.
Given that actual momentum

=\frac{h}{4{\pi }_m\times 0.05nm}

Substituting the values, we get

=\frac{6.626\times {10}^{-34}Js}{4\times 3.14\times 5.0\times {10}^{-11}m}

= 1.055 × 10^–24 kgms^–1
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

Q.61 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

1. n=4, l=2, m_l=-2, m_s=-1/2
2. n=3, l=2, m_l=1, m_s=+1/2
3. n=4, l=1, m_l=0, m_s=+1/2
4. n=3, l=2, m_l=-2, m_s=-1/2
5. n=3, l=1, m_l=-1, m_s=+1/2
6. n=4, l=1, m_l= 0, m_s=+1/2

Ans.

1. For n = 4 and l= 2, the orbital occupied is 4d.
2. For n = 3 and l= 2, the orbital occupied is 3d.
3. For n = 4 and l= 1, the orbital occupied is 4p.
4. For n=3 and l= 2, the orbital occupied is 3d.
5. For n=3 and l= 1, the orbital occupied is 3p.
6. For n=4 and l= 1, the orbital occupied is 4p.

Therefore, the increasing order of energies is

5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d)

Q.62 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

Ans.

In a multi-electron atom the effective nuclear charge experienced by an electron is dependent upon the distance between the nucleus and the orbital in which the electron is present. If the distance increases the effective nuclear charge decreases.

Among the given orbitals, the 4p orbital is the farthest from the nucleus, so it experiences the lowest attraction force by the nucleus. Again the electrons in the 4p orbital are shielded by electrons present in the 2p and 3p-orbitals along with the s-orbital. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge.

Q.63 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f,(iii) 3d and 3p

Ans.

Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. It is inversely proportional to the distance between the orbital and the nucleus.

1. The electrons present in the 2s orbital experience more effective nuclear charge than the 3s orbital because 2s orbital is closer to nucleus than 3s orbital.
2. 4d orbital experiences greater nuclear charge than 4f orbital, since 4d orbital is closer to nucleus.
3. 3p orbital is closer to the nucleus than 3d orbital, hence it will experience greater effective nuclear charge.

Q.64 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Ans.

Nuclear charge is directly proportional to the effective nuclear charge. The higher the atomic number higher will be the nuclear charge.

Silicon has 14 protons while aluminium has 13 protons in the nucleus. Hence, silicon has a larger nuclear charge of +14 than aluminium which has a nuclear charge of +13. The electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

Q.65 Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Ans.

(a) Phosphorus (P):

Atomic number = 15

The electronic configuration of P is: 1s²2s²2p⁶3s²3p³

The orbital picture of P can be represented as:

Phosphorus has three unpaired electrons.

(b) Silicon (Si):

Atomic number = 14

The electronic configuration of Si is: 1s²2s²2p⁶3s²3p²

The orbital picture of Si can be represented as:

Silicon has two unpaired electrons.

(c) Chromium (Cr):

Atomic number= 24

The electronic configuration of Cr is:

1s²2s²2p⁶3s²3p⁶4s¹3d⁵

The orbital picture of Cr can be represented as:

Chromium has six unpaired electrons.

(d) Iron (Fe):

Atomic number= 26

The electronic configuration is:

1s²2s²2p⁶3s²3p⁶4s²3d⁶

The orbital picture of Fe can be represented as:

Iron has four unpaired electrons.

(e) Krypton (Kr):

Atomic number= 36

The electronic configuration is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

The orbital picture of Kr can be represented as:

Krypton has no unpaired electrons, all orbitals are fully occupied.

Q.66 a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having m_s value of –1/2 for n= 4?

Ans.

(a) Given n=4,

For a given value of ‘n’, ‘l’ can have values from 0 to (n-1).

Thus, l= 0,1,2,3

Thus, four sub-shells are associated with n=4, which are s, p, d and f.

(b) The maximum number of electrons present in n^th shell =2n². For n=4, the maximum capacity of electrons

= 2×4² = 32.

Total number of orbitals = n² = 4²=16

If each orbital is taken fully, then it will have 1 electron with m_s value of –1/2.

Number of electrons with m_s value of (–1/2) is 16.

Q.67 If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ms^–1, calculate the de Broglie wavelength associated with it.

Ans.

According to de Broglie’s equation,

\lambda =h/mv

Where,
λ = wavelength associated with the electron
h= Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ

\lambda =\frac{6.626\times {10}^{-34}Js}{\left(9.10939\times {10}^{-31}kg\right)\left(2.19\times {10}^{6}m{s}^{-1}\right)}

= 3.32 x 10^-10 m= 3.32 x 10^-10 m x (100/100)
= 332 x 10^-12 m
λ = 332 pm
Wavelength associated with the electron = 332 pm