# NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

** NCERT solutions for class 12 physics chapter 11 Dual Nature of Radiation and Matter: ** Particle and the wave nature of matter are discussed in this chapter. In the solutions of NCERT class 12 physics chapter 11 dual nature of radiation and matter, you will study questions related to both particle and wave nature of light. Experimental study of the photoelectric effect is an important topic of NCERT for the CBSE board exam. CBSE NCERT solutions for class 12 physics chapter 11 dual nature of radiation and matter will be based on the understanding of such topics. Try to solve all the questions in NCERT yourself. If you are getting any doubt, you can refer to NCERT solutions. Understand all the graphs in NCERT class 12 physics chapter 11 dual nature of radiation and matter in-depth, which will help you to build an interest in the chapter.

The main formulas to be noted for NCERT solutions for class 12 physics of this chapter are given below.

Einstein’s photoelectric equation:

That is the maximum kinetic energy = energy of photon - work function

The De-Broglie wavelength of a charged particle is

is mass of the charged particle, K is the kinetic energy and V is the potential.

** NCERT solutions for class 12 physics chapter 11 dual nature of radiation and matter exercise **

** Q: 11.1 ** ** (a) ** Find the maximum frequency of X-rays produced by electrons.

** Answer: **

The X-Rays produced by electrons of 30 keV will have a maximum energy of 30 keV.

By relation,

** Q: 11.1 (b) ** Find the minimum wavelength of X-rays produced by electrons.

** Answer: **

From the relation , we have calculated the value of frequency in the previous questions, using that value and the following relation

** Answer: **

The energy of the incident photons is E is given by

Maximum Kinetic Energy is given by

** Answer: **

The stopping potential depends on the maximum Kinetic Energy of the emitted electrons. Since maximum Kinetic energy is equal to 0.34 eV, stopping potential is the maximum kinetic energy by charge equal to 0.34 V.

** Answer: **

The electrons with the maximum kinetic energy of 0.34 eV will have the maximum speed

** Answer: **

Since the photoelectric cut-off voltage is 1.5 V. The maximum Kinetic Energy (eV) of photoelectrons emitted would be 1.5 eV.

KE _{ max } =1.5 eV

KE _{ mac } =2.4 10 ^{ -19 } J

** Answer: **

The energy of photons is given by the relation

Momentum is given by De Broglie's Equation

The energy of the photons in the light beam is 3.14 10 ^{ -19 } J and the momentum of the photons is 1.046 10 ^{ -27 } kg m s ^{ -1 } .

** Q: 11.4 (b) ** ** ** Monochromatic light of wavelength is produced by a helium-neon laser. The power emitted is .

** Answer: **

Power of the light beam, P =9.42 mW

If n number of photons arrive at a target per second nE=P (E is the energy of one photon)

** Answer: **

Mass of Hydrogen Atom (m)=1.67 10 ^{ -27 } kg.

The speed at which hydrogen atom must travel to have momentum equal to that of the photons in the beam is v given by

** Answer: **

Average Energy(E) of the photons reaching the surface of the Earth is given by

Energy flux(I) reaching the Earth's surface=1.388 10 ^{ 3 } Wm ^{ -2 }

Number of photons(n) incident on Earth's surface per metre square is

** Q : 11.6 ** In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be . Calculate the value of Planck’s constant.

** Answer: **

The slope of the cut-off voltage versus frequency of incident light is given by h/e where h is Plank's constant and e is an electronic charge.

** Answer: **

The energy of a photon is given by

where h is the Planks constant, c is the speed of the light and lambda is the wavelength

** Answer: **

Power of the sodium lamp=100W

The rate at which photons are delivered to the sphere is given by

** Answer: **

Threshold frequency of the given metal( )=

The work function of the given metal is

The energy of the incident photons

Maximum Kinetic Energy of the ejected photoelectrons is

Therefore the cut off voltage is 2.025 eV

** Answer: **

The energy of photons having 330 nm is

Since this is less than the work function of the metal there will be no photoelectric emission.

** Q: 11.10 ** Light of frequency is incident on a metal surface. Electrons with a maximum speed of are ejected from the surface. What is the threshold frequency for photoemission of electrons?

** Answer: **

The energy of incident photons is E given by

Maximum Kinetic Energy of ejected electrons is

Work Function of the given metal is

The threshold frequency is therefore given by

** Answer: **

The energy of incident photons is given by

Cut-off potential is 0.38 eV

Work function is therefore, 2.54-0.38= 2.16 eV

** Q: 11.12 ** ** (a) ** Calculate the momentum of the electrons accelerated through a potential difference of .

** Answer: **

On being accelerated through a potential difference of 56 V the electrons would gain a certain Kinetic energy K.

The relation between Kinetic Energy and Momentum(p) is given by

** Q: 11.12 ** ** (b) ** ** ** Calculate the de Broglie wavelength of the electrons accelerated through a potential difference of .

** Answer: **

De Broglie wavelength is given by the De Broglie relation as

the wavelength is 0.164 nm

** Q: 11.13 ** ** (a) ** ** ** What is the momentum of an electron with kinetic energy of .

** Answer: **

The relation between momentum and kinetic energy is

** Q: 11.13 (b) ** What is the speed of an electron with kinetic energy of .

** Answer: **

The relation between speed and kinetic energy of a particle is

** Q: 11.13 ** ** (c) ** What is the de Broglie wavelength of an electron with kinetic energy of

** Answer: **

De Broglie wavelength is given by

The de Broglie wavelength associated with the electron is 0.112 nm

** Answer: **

The momentum of a particle with de Broglie wavelength of 589 nm is

The Kinetic Energy of an electron moving with above-mentioned momentum is

** Q: 11.14 (b) ** The wavelength of light from the spectral emission line of sodium is . Find the kinetic energy at which a neutron would have the same de Broglie wavelength.

** Answer: **

The momentum of the neutron would be the same as that of the electron.

The kinetic energy of neutron would be

** Q: 11.15 ** ** (a) ** What is the de Broglie wavelength of a bullet of mass travelling at the speed of .

** Answer: **

The momentum of the bullet is

De Broglie wavelength is

** Q: 11.15 ** ** (b) ** What is the de Broglie wavelength of a ball of mass moving at a speed of .

** Answer: **

The momentum of the ball is

De Broglie wavelength is

** Q: 11.15 ** ** (c) ** What is the de Broglie wavelength of a dust particle of mass drifting with a speed of ?

** Answer: **

The momentum of the dust particle is

De Broglie wavelength is

** Q: 11.16 ** ** (a) ** ** ** An electron and a photon each have a wavelength of . Find their momenta.

** Answer: **

Their momenta depend only on the de Broglie wavelength, therefore, it will be the same for both the electron and the photon

** Q: 11.16 ** ** (b) ** ** ** An electron and a photon each have a wavelength of . Find the energy of the photon.

** Answer: **

The energy of the photon is given by

h is the Planks constant, c is the speed of the light and lambda is the wavelength

** Q: 11.16 ** ** (c) ** ** ** An electron and a photon each have a wavelength of . Find the kinetic energy of electron.

** Answer: **

The kinetic energy of the electron is. In the below equation p is the momentum

** Q: 11.17 (a) ** For what kinetic energy of a neutron will the associated de Broglie wavelength be ?

** Answer: **

For the given wavelength momentum of the neutron will be p given by

The kinetic energy K would therefore be

** Q : 11.17 (b) ** Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of at .

** Answer: **

The kinetic energy of the neutron is

Where k Boltzmann's Constant is 1.38 10 ^{ -23 } J/K

The momentum of the neutron will be p

Associated De Broglie wavelength is

De Broglie wavelength of the neutron is 0.145 nm.

** Answer: **

For a photon we know that it's momentum (p) and Energy (E) are related by following equation

E=pc

we also know

Therefore the De Broglie wavelength is

The above de Broglie wavelength is equal to the wavelength of electromagnetic radiation.

** Q: 11.19 ** What is the de Broglie wavelength of a nitrogen molecule in air at ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen )

** Answer: **

Since the molecule is moving with the root-mean-square speed the kinetic energy K will be given by

K=3/2 kT where k is the Boltzmann's constant and T is the absolute Temperature

In the given case Kinetic Energy of a Nitrogen molecule will be

Mass of Nitrogen molecule = 2 14.0076 1.66 10 ^{ -27 } =4.65 10 ^{ -26 } kg

The momentum of the molecule is

Associated De Broglie wavelength is

The nitrogen molecule will have a De Broglie wavelength of 0.0275 nm.

**NCERT solutions for class 12 physics chapter 11 dual nature of radiation and matter additional exercise **

** Answer: **

The kinetic energy of an electron accelerated through Potential Difference V is K=eV where e the electronic charge.

Speed of the electrons after being accelerated through a potential difference of 500 V will be

Specific charge is e/m _{ e } =1.366 10 ^{ 11 } C/kg

** Answer: **

Using the same formula we get the speed of electrons to be 1.88 10 ^{ 9 } m/s. This is wrong because the speed of the electron is coming out to be more than the speed of light. This discrepancy is occurring because the electron will be travelling at very large speed and in such cases(relativistic) the mass of the object cannot be taken to be the same as the rest mass.

In such a case

where m is the relativistic mass, m _{ 0 } is the rest mass of the body, v is the very high speed at which the body is travelling and c is the speed of light.

** Answer: **

The force due to the magnetic field on the electron will be F _{ b } =evB (since the angle between the velocity and magnetic field is 90 ^{ o } )

This F _{ b } acts as the centripetal force required for circular motion. Therefore

** Answer: **

The formula used in (a) can not be used. As the electron would be travelling at a very high speed we can not take its mass to be equal to its rest mass as its motion won't be within the non-relativistic limits.

The value for the mass of the electron would get modified to

where m is the relativistic mass, m _{ 0 } is the rest mass of the body, v is the very high speed at which the body is travelling and c is the speed of light.

The radius of the circular path would be

** Answer: **

The kinetic energy of an electron after being accelerated through a potential difference of V volts is eV where e is the electronic charge.

The speed of the electron will become

Since the magnetic field curves, the path of the electron in circular orbit the electron's velocity must be perpendicular to the magnetic field.

The force due to the magnetic field is therefore F _{ b } =evB

This magnetic force acts as a centripetal force. Therefore

** Answer: **

The wavelength of photons with maximum energy=0.45

Energy of the photons is

** Q: 11.23 (b) ** From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

** Answer: **

In such a tube where X-ray of energy 27.6 keV is to be produced the electrons should be having an energy about the same value and therefore accelerating voltage should be of order 30 KeV.

** Answer: **

The total energy of 2 rays=10.2 BeV

The average energy of 1 ray, E=5.1 BeV

The wavelength of the gamma-ray is given by

** Q: 11.25 (a) ** Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. The number of photons emitted per second by a Medium wave transmitter of power, emitting radiowaves of wavelength .

** Answer: **

The power emitted by the transmitter(P) =10kW

Wavelengths of photons being emiited=500 m

The energy of one photon is E

Number of photons emitted per second(n) is given by

** Answer: **

The minimum perceivable intensity of white light(I)=10 ^{ -10 } Wm ^{ -2 }

Area of the pupil(A)=0.4 cm ^{ 2 } =4 10 ^{ -5 } m ^{ 2 }

Power of light falling on our eyes at minimum perceivable intensity is P

P=IA

P=10 ^{ -10 } 4 10 ^{ -5 }

P=4 10 ^{ -15 } W

The average frequency of white light( )=6 10 ^{ 14 } ^{ } Hz

The average energy of a photon in white light is

Number of photons reaching our eyes is n

** Answer: **

The energy of the incident photons is E given by

Since stopping potential is -1.3 V work function is

The energy of photons which red light consists of is E _{ R }

Since the energy of the photons which red light consists of have less energy than the work function, there will be no photoelectric emission when they are incident.

** Answer: **

The wavelength of photons emitted by the neon lamp=640.2 nm

The energy of photons emitted by the neon lamp is E given by

Stopping potential is 0.54 V

Work function is therefore

The wavelength of photons emitted by the iron source=427.2 nm

The energy of photons emitted by the ion source is

New stopping voltage is

The stopping voltages, respectively, were measured to be

** Answer: **

where V is stopping potential, h is planks constant, e is electronic charge, is frequency of incident photons and is work function of metal in electron Volts.

To calculate the planks constant from the above date we plot the stopping potential vs frequency graph

The plot we get is

From the above figure, we can see that the curve is almost a straight line.

The slope of the above graph will give the Plank's constant divided by the electronic charge. Planks constant calculated from the above chart is

Planks constant calculated from the above chart is therefore

** Q: 11.29 ** ** ** The work function for the following metals is given:

** Answer: **

The wavelength of the incident photons=

The energy of the incident photons is

Mo and Ni will not give photoelectric emission for radiation of wavelength from a .

If the laser is brought nearer no change will be there in case of Mo and Ni although there will be more photoelectrons in case of Na and K.

** Answer: **

Intensity of Incident light(I) =

The surface area of the sodium photocell (A)=2 cm ^{ 2 } = 2 10 ^{ -4 } m ^{ 2 }

The rate at which energy falls on the photo cell=IA=2 10 ^{ - } ^{ 9 } W

The rate at which each of the 5 surfaces absorbs energy= IA/5=4 10 ^{ -10 } W

Effective atomic area of a sodium atom (A')= 10 ^{ -20 } m ^{ 2 }

The rate at which each sodium atom absorbs energy is R given by

The time required for photoelectric emission is

** Answer: **

According to De Broglie's equation

The kinetic energy of an electron with De Broglie wavelength is given by

The kinetic energy of photon having wavelength is

Therefore for the given wavelength, a photon has much higher energy than an electron.

** Answer: **

Kinetic energy of the neutron(K)=150eV

De Broglie wavelength associated with the neutron is

Since an electron beam with the same energy has a wavelength much larger than the above-calculated wavelength of the neutron, a neutron beam of this energy is not suitable for crystal diffraction as the wavelength of the neutron is not of the order of the dimension of interatomic spacing.

** Answer: **

Absolute temperature = 273+27=300K

Boltzmann's Constant=1.38 10 ^{ -23 } J/mol/K

The de Broglie wavelength associated with the neutron is

Since this wavelength is comparable to the order of interatomic spacing of a crystal it can be used for diffraction experiments. The neutron beam is to be thermalised so that its de Broglie wavelength attains a value such that it becomes suitable for the crystal diffraction experiments.

** Answer: **

The potential difference through which electrons are accelerated(V)=50kV.

Kinetic energy(K) of the electrons would be eV where e is the electronic charge

The De Broglie wavelength associated with the electrons is

The wavelength of yellow light = 5.9 10 ^{ -7 } m

The calculated De Broglie wavelength of the electron microscope is about 10 ^{ 5 } more than that of yellow light and since resolving power is inversely proportional to the wavelength the resolving power of electron microscope is roughly 10 ^{ 5 } times than that of an optical microscope.

(Rest mass energy of electron .)

** Answer: **

Rest mass of the electron

Momentum

using the relativistic formula for energy

** Q: 11.35 ** Find the typical de Broglie wavelength associated with a atom in helium gas at room temperature ( ) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

** Answer: **

The kinetic energy K of a He atom is given by

m _{ He } i.e. mass of one atom of He can be calculated as follows

(N _{ A } is the Avogadro's Number)

De Broglie wavelength is given by

The mean separation between two atoms is given by the relation

From the ideal gas equation we have

The mean separation is therefore

The mean separation is greater than the de Broglie wavelength.

** Answer: **

The de Broglie wavelength associated with the electrons is

The de Broglie wavelength of the electrons is comparable to the mean separation between two electrons.

** ** Answer the following questions:

** Answer: **

Quarks are thought to be tight within a proton or neutron by forces which grow tough if one tries to pull them apart. That is event though fractional charges may exist in nature, the observable charges are still integral multiples of the charge of the electron

** ** Answer the following questions:

** Q: 11.37 (b) ** What is so special about the combination ? Why do we not simply talk of and separately?

** Answer: **

The speed of a charged particle is given by the relations

or

As we can see the speed depends on the ratio e/m it is of such huge importance.

Answer the following questions:

** Q: 11.37 (c) ** Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

** Answer: **

At ordinary pressure due to a large number of collisions among themselves, the gases have no chance of reaching the electrodes while at very low pressure these collisions decrease exponentially and the gas molecules have a chance of reaching the respective electrodes and therefore are capable of conducting electricity.

Answer the following questions:

** Answer: **

The work function is defined as the minimum energy below which an electron will never be ejected from the metal. But when photons with high energy are incident it is possible that electrons from different orbits get ejected and would, therefore, come out of the atom with different kinetic energies.

Answer the following questions:

** Answer: **

The absolute energy has no significance because of the reference point being arbitrary and thus the inclusion of an arbitrary constant rendering the value of and . to have no physical significance as such.

The group speed is defined as

Due to the significance of the group speed the absolute value of wavelength has physical significance.

## ** ****NCERT solutions for class 12 physics chapter wise **

** NCERT solutions subject wise **

## ** Importance of NCERT solutions for class 12 physics chapter 11 dual nature of radiation and matter: **

- The chapter dual nature of radiation and matter comes under modern physics part of class 12 which is important for competitive exams like NEET and JEE Mains.
- As far as CBSE board exam is considered, 3 to 5 marks questions are expected from this chapter. In the year 2019 CBSE board exam, three marks questions were asked from this chapter.
- The CBSE NCERT solutions for class 12 physics chapter 11 dual nature of radiation and matter will help to score well in the board and competitive exams.

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

**Question: **What is the weightage of chapter dual nature of matter and radiation for CBSE board exam

**Answer: **

On an average 5 marks, questions can be expected from the NCERT class 12 chapter 11

**Question: **Which topics are important from class 12 physics chapter dual nature of matter and radiation

**Answer: **

All the topics of the chapter are important. Understand all the graphs mentioned in the NCERT book

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