CBSE Important Questions Class 12 Physics Chapter 11

Important Questions for CBSE Class 12 Physics Chapter 11 – Dual Nature of Radiation and Matter

With these important questions for Class 12 Physics chapter 11, you will be introduced to the topic of the dual nature of radiation and matter. These chapter 11 Class 12 Physics important questions contain many important questions and formulas that students should study before exams. These Physics Class 12 chapter 11 important questions contain questions based on the NCERT books and prescribed CBSE syllabus.

After studying these Class 12 Physics chapter 11 important questions, students will be able to solve CBSE past years’ question papers. In addition, these CBSE revision notes also contain CBSE extra questions for students to try to test their grasp of the concepts.

CBSE Class 12 Physics Chapter-11 Important Questions

Study Important Questions for Class 12 Physics Chapter– Dual Nature of Radiation and Matter

Short 1 marks answer type questions

2. The photoelectron has a maximum kinetic energy of 2.8 eV. What is the stopping potential worth?

Ans. 12mv²=eVo=2.8eV

Vo=2.8V

1. The work functions of two metals denoted as A and B are 4 eV and 10 eV, respectively. Which metal has a higher wavelength threshold?

Ans. The work function has an inverse relationship with the threshold wavelength. Metal A, therefore, has a longer threshold wavelength.

1. Two photosensitive materials with work functions W1 and W2 (where W1 > W2) are exposed to ultraviolet light. Which scenario will result in higher kinetic energy for the electrons being released and why?

Ans. The photoelectron’s kinetic energy can be given as 12mv²=hv-W. The more energy they have, the lower the work function must be for a given frequency. The metal with work function W2 will have more kinetic energy since W2 is greater than W1.

1. When the distance between the light source and the photocell’s cathode is doubled, how does the stopping potential applied to the cell change?

Ans. The light source’s intensity, which varies as the distance from its changes, has no bearing on the stopping potential.

1. 1.5 V serves as the stopping potential in a photoelectric effect experiment. What is the photoelectrons’ highest possible kinetic energy?

Ans. The electron’s K.E. is e– =1.5 eV.

1. In the photoelectric effect’s context, define “stopping potential”.

Ans. For a specific frequency of incident radiation, the cut-off or stopping potential is the value of the retarding potential at which the photoelectric current zeroes out.

1. Explain the de Broglie hypothesis.

Ans. According to the de Broglie hypothesis, a mass particle moving at a particular speed v must be connected to a matter waver at a specific wavelength X.

=hp=hmv

1. Why is it impossible to produce photoelectric emission at all frequencies?

Ans. Only when the incident photon’s energy (hv) exceeds the metal’s work function (0 = hv0) is photoelectric emission conceivable. Therefore, the incident radiation’s frequency v must be higher than the threshold frequency v0.

1. Is it more difficult to release a free electron from copper than sodium, and why?

Ans. Since 0=hc0

Here, 0 represents the threshold wavelength

Since 0Na>0Cu

∴ Copper has a higher work function, making it more challenging to remove a free electron from it than sodium.

1. Identify the relationship between the threshold frequency and the incident radiation frequency that describes the maximum kinetic energy of the electrons that can be emitted from a metal surface.

Ans.  K.Emax=hv-w

Here, W represents the work function or the energy of the threshold, which depends on the threshold frequency of v0/

w=hv0

Or 12mv²max=hv-hv0

Or

hv=12mv²max+hv0

1. Assume that the radiation hitting a metal plate has a frequency higher than its threshold frequency. What would happen if the incidence of radiation was doubled?

(1) Electron kinetic energy

(2) Photoelectric current

Ans. (1) When the frequency of incident radiation doubles, the kinetic energy also doubles.

(2) The number of photoelectrons, or photoelectronic current, won’t change if the frequency of the incident radiation is doubled.

1. In one experiment, the photoelectric cut-off voltage is 1.5 V. What is the highest kinetic energy of photons released?

Ans. The information given is that

The photoelectric cut-off voltage is V0=1.5 V

The highest K.E. of the emitted photoelectrons can be represented as 

Ke=eV0

Where, 

e=charge on an electron= 1.610-19 C

∴ Ke=1.610-19 1.5 

=2.410-19J 

Hence, the maximum K.E. of the emitted photoelectrons in the mentioned experiment will be 

2.410-19J.

1. Through a 64-volt potential difference, an electron is accelerated. What is the corresponding de Broglie wavelength? What region of the electromagnetic spectrum does this wavelength value fit into?

Ans. As per the de Broglie wavelength 

=1.227Vnm=1.22764= 1.2278= 0.1533 nm

This wavelength is often associated with x-rays. 

Short 3 marks question answers

1. Write the photoelectric equation given by Einstein. List the three key characteristics of the photoelectric effect that may be explained using the aforementioned equation.

Ans. The photoelectric equation given by Einstein is Kmax=hv-0

We discover that Kmax only depends linearly on V. It is unaffected by the radiation’s strength.

Since Kmax must always be positive

hv>0 v>v0              (∴0=hv0)

So, if the work function is greater(0), the minimum frequency or the threshold frequency required for the purpose of emitting photoelectrons will also be higher.

Also, the number of photoelectrons increases as the number of energy quanta increases. As a result, photoelectric current and intensity are inversely related.

1. Using de Broglie’s hypothesis, derive Bohr’s quantization condition for the angular momentum of the electron orbiting in the hydrogen atom.

Ans. According to Broglie’s hypothesis,

We get =hp=hmvn

From Bohr’s postulate,

2rn=n                     (n=1,2,3)

∴ 2rn=nhmvn

mvnrn=nh2

∴ Angular momentum= (mvnrn) =nh2

4. A certain metal’s work function is 4.2 eV. Will this metal emit photoelectrically for incident light with a 330 nm wavelength?

Ans. It will not.

We are given that,

Metal’s work function, 0=4.2eV

Electron’s charge, e=1.610-19C

Planck’s constant, h= 6.62610-34Js

The wavelength of the incident radiation, =330m = 33010-9m

Speed of light, c=3108 m/s

The incident photon’s energy can be given as

E=hc

=6.62610-34310833010-9

=6.010-191.610-19= 3.76eV

We can observe that the incident radiation energy is lower as compared to the metal’s work function. Therefore, there will be no photoelectric emission.

1. Define the photoelectric effect’s threshold wavelength

Ans. Threshold wavelength refers to the maximum wavelength of radiation that is required to cause photoelectric emission. 

1. 1ev is a metal’s photoelectric work function. It receives wavelength light of 3000A°.

What is the affected photoelectron’s velocity?

Ans. =hm or m or m=h

K.E(E’)= h22m2           

K.E=P22m

(K.E.) electron(K.E.) alpha= mme      (∴=hP is same)  

1. Find the following:

(a) The highest frequency, and

(b) The highest wavelength of X-rays produced by 30 kV electrons.

Ans. The electron’s potential, V=30 kV= 3104 V

Hence, the electron’s energy, E=3104 eV

Here, e is the change present on the electron= 1.610-19 C

(a)  The maximum frequency that is produced by the x-rays=v

The electron’s energy can be given by the following relation:

E=hv

Here, H is the Planck constant that is 6.62610-34 Js

∴ v= 7.241018 Hz

(b) The minimum wavelength that can be produced by the x-rays can be given as

=cv

= 31087.241018=4.1410-11m=0.0414 nm.

Therefore, the minimum wavelength that can be produced by the x-rays is 0.0414 nm.

1. The amount of solar energy that touches the earth’s surface is 1.388103 W/m². How many photons are incident on the Earth each second per square metre (almost)? Assume that the photons present in the sunlight have an approximate wavelength of 550 nm.

Ans. The sunlight’s energy flux that reaches the earth’s surface is 1.388103 W/m².

Therefore, the sunlight’s power per square metre is 

P=1.388103 W/m²

The speed of light, C= 3108 m/s

Planck’s constant, h=6.62610-34 Js

The average wavelength of photons that exist in the sunlight, =550 nm

=55010-9 m

The number of photons per square metre that is incident on the earth per second= n

Therefore, we can write the equation of power as 

P=nE

∴ n=PE=Phc

=1.38810355010-96.62610-343108= 3.841021 photons/m²/S

Hence, the photons that are incident per square metre on earth every second is 3.841021.

1. List two characteristics of photons. Why aren’t all photoelectrons produced by a monochromatic photon incident on a photosensitive surface at the same energy? Explicitly justify your response.

Ans. Two features of photons are:

• The photon has no electrical charge.
• The energy of a photon is hv.

All photoelectrons do not exit from a monochromatic radiation incident on a photosensitive surface with the same energy as, in addition to the work required to remove electrons from the surface, various (emitted) photoelectrons require varying amounts of work in order to reach the surface.

1. The following graph displays the relationship between stopping potential V0 and incident radiation frequency v for two photosensitive metals, X and Y:

(i) What metal has the longer threshold wavelength? Give reason.

(ii) Identify the metal that emits electrons with higher kinetic energy for the same wavelength of incident radiation and explain why.

(iii) How would the kinetic energy of electrons emitted from metal X vary if the distance between the light source and it is cut in half? Give reason.

Ans. 

(i) =Cv

As (v0)X < (v0)Y   ∴   (0)X > (0)Y

∴ Metal ‘X’ has the larger threshold wavelength

(ii) As per Einstien’s photoelectric equation:

hC=hC0 + K.E. of photoelectron

L.H.S. is constant for the same of incident radiation. As a result, metal X with a higher value of 0 will emit photoelectrons with a greater K.E.

(ii) Energy in motion won’t alter. Only the intensity of light varies as the distance shrinks; the frequency stays the same. K.E. depends on the frequency of photoelectrons that are released.

1. Answer the following:

(i) Why can’t the photoelectric effect be described using the fact of the wave nature of light? 

(ii) Write down the fundamental characteristics of electromagnetic radiation’s photon model, which forms the basis of Einstein’s photoelectric equation.

Ans. (i)  Although it should, experimental data does not show that the maximum kinetic energy of the released electron is directly proportional to the intensity of incident radiations. According to wave theory, the maximal kinetic energy of the released electrons should not rely on the incident frequency, but this is not the case.

The threshold frequency should not exist, according to wave theory. If the light’s intensity is high enough for electrons to escape, the light of all frequencies should do so.

The photoelectric effect shouldn’t happen instantly, according to wave theory. The wave’s energy cannot be transferred to a specific electron; rather, it is spread to all of the electrons in the lighted region as a whole. Therefore, a delay between the incidence of radiation and the emission of electrons is necessary.

(ii) Radiation behaves as though it were composed of photon-like particles. Every photon has momentum p = h/ and energy E = hv. The quantity of photons that fall on the surface each second provides an understanding of radiation intensity. Photon energy is intensity-independent and only dependent on frequency.

An electron and a photon colliding one-to-one can be used to explain the photoelectric effect.

A photon of a certain frequency strikes a metal surface, using some of its energy to overcome the work function and some of it to impart kinetic energy, so KE = h(v – v0).

Long answer questions

1. Electrons are accelerated by a voltage of 50 kV in an electron microscope. Find the electrons’ corresponding de-Broglie wavelength. Assuming other criteria like numerical aperture etc., are the same, how is the electron microscope’s resolving power comparable to that of an optical microscope that employs yellow light?

Ans. we are given that: V=50 kV= 50103 volts

The kinetic energy of the electron,

EK=5010³ eV

= 5010³1.610-19 J

=501.610-16

De Broglie wavelength of an electron is 

=h2mEk

=(6.6310-34)29.1110-31501.610-16

= 5.5 10-12 m

Since resolving power (R.P.) is its wavelength are inversely proportional, R.P. is equal to wavelength X = 5.910-7 m for yellow light. 105 times more than that of an optical microscope for an electron microscope.

1. The slope of the cut-off voltage vs frequency of incident light is discovered to be V s in a photoelectric effect experiment. Determine the Planck constant’s value.

Ans. The slope of the cut-off voltage (v) versus frequency () of incident light is given as

V= 4.1210-15 Vs

V that is related to the frequency by the equation:

hv=eV

Here, e is the charge present on an electron, i.e., 1.610-19 C, and h is Planck’s constant

∴  h=eV

= 1.610-19 4.1210-15 = 6.59210-34 js

Hence, 6.59210-34  will be Planck’s constant.  

632. Using a helium-neon laser, monochromatic light with a wavelength of 632.8 nm is made. 9.42 mW are released as energy.

(a) Determine each photon’s energy and momentum within the light beam.

(b) How many photons arrive at a target exposed to this beam on average each second? (take the assumption that the beam has a uniform cross-section smaller than the target region), and

(c) How fast must an atom of hydrogen move in order to match the momentum of a photon?

Ans. The monochromatic light’s wavelength, =632.8 nm= 632.810-9 m

Power that is emitted by the laser, P= 9.42 mW= 9.4210-3 W

The Planck’s constant, h= 6.62610-34 Js

Light’s speed= 3108 m/s

Hydrogen atom’s mass, m=1.6610-27 kg

(a) The energy of a single photon can be given as

E=hc

=6.62610-343103632.810-9= 3.14110-19 J

The momentum of a single photon can be given as:

P=h

=6.62610-34632.8= 1.04710-27 kg ms-1

(b) Number of photons that arrive at a target each second after being exposed to the beam=n

Assume the uniform cross-section of the beam is smaller than the target region. Consequently, the power equation can be expressed as:

P=h

∴ n=PE

=9.4210-33.14110-19 31015 photon/s

(c) Hydrogen atom’s momentum and photon’s momentum are the same

p= 1.04710-27 kg ms-1

Momentum can be given as:

p=mv

Here, v denotes hydrogen atom’s speed.

∴ = pm

=1.04710-271.6610-27=0.621 m/s

1. List the key photon characteristics that Einstein used to develop his photoelectric equation.

(a) Use this equation to clarify the idea of

• Threshold frequency and

• Stopping potential.

Ans: (a) Some important properties of photons are as follows:

• When radiation interacts with matter, it behaves as though it were formed of particles called photons.
• Each photon has speed c, the speed of light, as well as energy E (= hv), momentum p (= hv/c), and energy.
• Regardless of the radiation’s intensity, all photons of light with a certain frequency v, or wavelength , possess the same energy E (= hv/c = h/) and momentum p (= hv/c = h/). The number of photons per second that cross a particular region is only increased by raising the intensity of light of a certain wavelength, with each photon having the same energy. As a result, photon energy is independent of radiation intensity.
• Since photons are electrically neutral, magnetic and electric forces have no effect on them.
• The complete energy and total momentum are conserved in photon-particle collisions (like photon-electron collisions). In a collision, the quantity of photons might not be conserved. Either a new photon will be formed, or the photon may be absorbed.

(b) The photoelectric equation of Einstein is

Kmax=hv-0    

Since Kmax must be non-negative, equation (i) implies that photoelectric emission is possible only if 

hv>0  or v>v0, where v0=0h

This equation demonstrates that the threshold frequency v0 required to generate photoelectrons increases with increasing work function 0.

Therefore, no matter how powerful the incident radiation is or how long it may remain on the surface, below a certain frequency, v0 (=0/h, no photoelectric emission is theoretically possible from on the metal surface.

Stopping potential is the smallest amount of negative potential (v0) that should be given to the anode in a photocell in order to make the photoelectric current zero. 

∴ Kmax=12mv²max=eV0

eV0=hv-0  for v v0

or V0=hev-0e

The stopping potential will be V0.

1. List three features of the photoelectric effect that can only be explained by Einstein’s equation and cannot be explained using the wave theory of light.

Ans. Here are three features of the photoelectric effect. 

• The experimental data does not show that the maximum kinetic energy of the released electron is directly proportional to the strength of incident radiations, but it should. According to wave theory, the maximal kinetic energy of the released electrons should not rely on the incident frequency, but this is not the case.
• The threshold frequency should not exist, according to wave theory. If the light’s intensity is high enough for electrons to escape, the light of all frequencies should do so.
• The photoelectric effect shouldn’t happen instantly, according to wave theory. The wave’s energy cannot be transferred to a specific electron; rather, it is spread to all of the electrons in the lighted region as a whole. Therefore, a delay between the incidence of radiation and the emission of electrons is necessary.

1. What will be the de Broglie wavelength in the following cases:

(a) a bullet having a mass 0.040 kg and is travelling at 1.0 km/s speed.

(b) a ball having a mass of 0.060 kg that is moving at 1.0 m/s speed.

(c) a dust particle having a mass 1.010-9 that drifts at a speed of 2.2 m/s

Ans. (a) The bullets’ mass, m=0.040 kg

The bullet’s speed, = 1.0 km/s= 1000 m/s

Planck’s constant, h=6.610-34 Js

The bullet’s de Broglie wavelength can be given by the following relation

=hmv

=6.610-340.0401000=1.6510-35 m

(b)The ball’s mass, m=0.060 kg

The balls speed, = 1.0 m/s

The ball’s De Broglie wavelength can be given by the following relation

  =hmv

= 6.610-340.0601=1.110-32 m

(c) The dust particle’s mass, m=110-9 kg

The dust particle’s speed, =2.2 m/s

The dust particle’s De Broglie wavelength can be given by the following relation

=hmv

=6.610-342.210-9=3.010-25 m²

1. Show that the de Broglie wavelength of an object’s quantum (photon) corresponds to the wavelength of electromagnetic radiation.  

Ans. The speed of photon and its energy (hv) can be given as

p=hvc=h

=hp

Here, is the electromagnetic radiation’s wavelength, c is the speed of light, and h is the Planck’s constant.

The photon’s de Broglie wavelength can be given as

=hmv

But p=mv

∴  =hp

Here, m is the photon’s mass, and v is the photon’s velocity.

Therefore, we can infer from (i) and (ii) equations that electromagnetic radiation’s wavelength and the photon’s de Broglie wavelength are equal. 

1. Answer the following questions. 

(i) Name two key components of Einstein’s photoelectric equation.

(ii) Two photosensitive surfaces P and Q are incidents with radiation at a frequency of 1015 Hz. Surface P shows no photoemission. Surface Q produces photoemission, but the photoelectrons have no kinetic energy. Find the work function value for surface Q and explain these observations.

Ans. (i) Defining characteristics of the photoelectric effect:

• Radiation behaves as though it were composed of photon-like particles. Every photon has momentum p = h/ and energy E = hv.
• The quantity of photons that fall on the surface each second provides an understanding of radiation intensity. Photon power is independent of intensity and simply depends on frequency.
• The one-to-one collision between an electron and a photon can be used to explain the photoelectric effect.
• A photon of a certain frequency strikes a metal surface, using some of its energy to overcome the work function and some of it to impart kinetic energy, resulting in KE = h. (v – v0)

(ii) Since P does not emit any photoelectric energy, the frequency of incident radiation (1015 Hz) must be below P’s threshold frequency (v0)p. Q produces photoemission, but the photoelectrons’ kinetic energy is 0. This suggests that the incident radiation frequency is exactly equal to the Q’s threshold frequency.

For Q, work function 0=hv0,

=(6.610-34)(1015)(1.610-19) eV=4.125 eV.

1. Give reasons for the following and explain:

(a) An increase in incident radiation intensity causes an increase in photoelectric current in a photocell.

(b) For a specific photosensitive surface, the stopping potential (V0) is in accordance with the frequency (v) of the incident radiation, with the slope staying constant across surfaces.

(c) The photoelectrons’ maximum kinetic energy is unaffected by the strength of the incident radiation.

Ans. (a) A photoelectron may be released as a result of a photon collision (above the threshold frequency). The quantity of photons increases with intensity. Consequently, the current grows.

(b) We have eVs=h(–0)

∴ Vs=he()+–kv0e

∴ The graph of Vs with b is a straight line, and the slope he is a constant. 

(c) since the maximum kinetic energy for different surfaces is given by

(K.E.)max= h(–0)

Therefore, maximum kinetic energy is dependent on frequency rather than incident radiation intensity. 

1. Answer the following: 

(a) List three photoelectric effect characteristics that have been observed but cannot be explained by the light wave hypothesis. How these qualities can be adequately described by using Einstein’s photoelectric equation?

(b) For two photosensitive materials, M1 and M2, the stopping potential (v0) vs frequency (v) of the incoming radiation are plotted in the figure. 

(i) Explain why the slope of both lines is the same.

(ii) Which material emits electrons with higher kinetic energy for the same incident radiation frequency?

Ans. Here are three photoelectric effect characteristics: 

• The maximum kinetic energy of the liberated electron does not appear to be directly proportional to the strength of the incident radiations, despite the fact that this should be the case according to experimental data.
• The threshold frequency should not exist, according to wave theory. If the light’s intensity is high enough for electrons to escape, the light of all frequencies should do so.
• The photoelectric effect shouldn’t happen instantly, according to wave theory. The wave’s energy cannot be transferred to a specific electron; rather, it is spread to all of the electrons in the lighted region as a whole. It follows that there must be a lag between the incidence of radiation and the emission of electrons.

 (b) Planck’s constant (h) = 6.6210-34JS is the universal constant shown by both lines’ same slope (V0/v).

M1 will have more kinetic energy at the same frequency of incident radiations because M1 material has a higher V0 value. Drawing a vertical line with the same frequency and having M1 and M2 connect at various points can clearly demonstrate it (V0 for M1, is higher)

1. Answer the following:

(a) The quantity of photons released each second by a 10 kW medium wave transmitter with a 500 m radio wave emission wavelength.

(b) The quantity of photons entering our eye’s pupil per second that corresponds to

the lowest white light intensity that humans are able to perceive.

Assume that the pupil is 0.4 cm2 in size and that white light has an average frequency to be approximately 61014 Hz.

Ans. (a) The power of the medium wave transmitter, P=10 kW= 104 W= 104 J/s

Therefore, the energy that the transmitter emits per second is E=104

The radio wave’s wavelength is, = 500 m

The wave’s energy is 

E1=hc

Here, H is the Planck’s constant that equals to 6.610-34 Js, and c represents the speed of light that equals to 3108 m/s

∴ E1=6.610-343108500= 3.9610-28 J

Let the number of photons that the transmitter emits to be n

∴ nE1=E

n=EE1

=1043.9610-28= 2.5251031

31031

A radio photon has very little energy (E1), but a radio wave emits a high number of photons (n) each second.

The total energy of a radio wave can be thought of as continuous, and the existence of a minimum quantum of energy can be disregarded.

(b) The light’s intensity that the human eye perceives, I=10-10 W m-2

The pupil’s area, A= 0.4 cm²= 0.410= 4m²

The white light’s frequency, =61014 Hz

The energy that the photon emits can be given as

E=hv

Here, h is the Planck’s constant= 6.610-34 Js

∴ E= 6.610-34 61014 

=3.9610-19 J

Let the total amount of photon that is falling per second and per pupil’s unit area to be n

The energy of the n photons per unit can be given as 

E=n3.9610-19 Js-1m-2

The intensity of light is defined as the energy per unit area per second.

∴ E=I

n3.9610-19=10-10

n=10-103.9610-19s

=2.52108m²s-1

The total photons entering per second in the pupil can be given as 

nA=nA

=2.521080.410-4

=1.008104 s-1

While this amount is not as big as compared to the figure in the first problem, it is still large enough to be undetected by the human eye.

Dual Nature of Radiation and Matter

These chapter 11 class 12 physics important questions have formulas that you should learn as they will prepare you for your exams. Here are some topics that you will learn in this chapter.

 Photoelectric effect: It occurs when a light beam of the right frequency is incident on a metal surface and causes photoelectron emission.

Hertz’s observation: During his study of electromagnetic waves, Heinrich Hertz made the discovery of photoelectric emission in 1887. When the emitter plate was lit by ultraviolet light from an arc lamp, the electromagnetic waves produced by the spark over the detector loop in his experiment were amplified.

Leonard’s observation: Lenard discovered that current flows when ultraviolet light is permitted to hit the emitter plate of a two-electrode evacuated glass tube. The current flows ceased as soon as the UV radiations were cut off.

Some terms related to the photoelectric effect

(i) A free electron: The electrons in the outer shells of metals, known as valence electrons, are only weakly connected to the atoms; as a result, they are free to move around freely on the metal surface but are unable to penetrate it. These are referred to as free electrons.

(ii) Electron emission: The emission of electrons from a metal’s surface is a phenomenon known as electron emission.

(iii) Photoelectric emission: It is a process wherein metal surfaces emit electrons when appropriate frequency light radiation strikes them.

(iv) Work function:  The work function of a metal is the bare minimum of energy needed to simply evict an electron from its outermost surface.

(v) Cut-off potential: Cut-off or stopping potential is the lowest negative (retarding) potential V0 that is applied to a plate for a certain frequency of incident radiation at which the photoelectric current is zero.

(vi) Cut-off frequency: The threshold frequency or cut-off frequency of a material is the light’s lowest frequency that can cause that substance to emit photoelectrons.

(vii) Cut-off wavelength: The threshold wavelength or cut-off wavelength of a material refers to the highest wavelength of light from which it can emit photoelectrons.

Effect of intensity of light on photocurrent: The photoelectric current grows linearly as incident light intensity increases for a fixed frequency of radiation.

Effect of potential on the photoelectric current: The photoelectric current rises with rising potential supplied to the collector for a certain frequency and intensity of incident light. Saturation current is the greatest current that exists when all of the photoelectrons have reached plate A.

Effect of frequency of incident radiation: We absorb radiation with the same intensity across all frequencies. We examine the variation in photoelectric current for each radiation in relation to the potential difference between the plates.

Einstein photoelectric equation: The energy quantum of radiation, Kmax=hv-0 when hv is the energy of a photon and is the work function.

Laws of photoelectric emission

(i) The number of photoelectrons expelled per second for a particular material and a given frequency of incident radiation is exactly proportional to the intensity of the incident light.

(ii) It is discovered that the saturation current for a particular material and frequency of incident radiation is proportional to the intensity of the incident radiation. In contrast, the stopping potential is independent of the intensity.

(iii) There is a specific minimum frequency of incident light for a given material below which photoemissions of photoelectrons do not occur. The term “threshold frequency” refers to this frequency.

The maximum kinetic energy of the generated photoelectron or equivalent stopping potential depends only on the frequency (or wavelength) of the incident light above the threshold frequency and is independent of the intensity of the input light.

(iv) The photoelectric emission takes place instantly. Less than even 10-9 seconds separate the incident of radiations from photoelectron emission.