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 this chapter 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 solutionsUnderstand 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 chapter 11 dual nature of radiation and matter 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

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

By relation,

$\\eV_{0}=h\nu \\ \nu =\frac{eV_{0}}{h}\\ \nu =\frac{1.6\times 10^{-19}\times 30\times 10^{3}}{6.62\times 10^{-34}}\\ \nu =7.25\times 10^{18}\ Hz$

From the relation   $eV_{0}=h\nu$, we have calculated the value of frequency in the previous questions, using that value and the following relation

$\\\lambda =\frac{c}{\nu }\\ \lambda =\frac{3\times 10^{8}}{7.25\times 10^{18}}\\ \lambda =0.04\ nm$

The energy of the incident photons is E is given by

$\\E=h\nu \\ E=\frac{6.62\times 10^{-34}\times 6\times 10^{14}}{1.6\times 10^{-19}}\\ E=2.48\ eV$

Maximum Kinetic Energy is given by

$\\KE_{max}=E-\phi _{0}\\ KE_{max}=2.48-2.14\\ KE_{max}=0.34\ eV$

Q: 11.2 (b) The work function of caesium metal is $\dpi{100} 2.14\hspace{1mm}eV$ When light of frequency $\dpi{100} 6\times 10^1^4\hspace{1mm}Hz$  is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential

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.

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

$\\KE_{max}=0.34\ eV\\ KE_{max}=5.44\times 10^{-20}\ J\\ v_{max}=\sqrt{\frac{2KE_{max}}{m}}\\v_{max}=\sqrt{\frac{2\times 5.44\times 10^{-20}}{9.1\times 10^{-31}}}\\v_{max}=3.44\times 10^{5}\ ms^{-1}$

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

KEmax=1.5 eV

KEmac=2.4$\times$10-19 J

The energy of photons is given by the  relation

$\\E=h\nu \\ E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{632\times 10^{-9}}\\ E=3.14\times 10^{-19}\ J$

Momentum is given by De Broglie's Equation

$\\p=\frac{h}{\lambda }\\ p=\frac{6.62\times 10^{-34}}{632.8\times 10^{-9}}\\ p=1.046\times 10^{-27}\ kg\ m\ s^{-1}$

The energy of the photons in the light beam is 3.14$\times$10-19 J and the momentum of the photons is 1.046$\times$10-27 kg m s-1.

How many photons per second, on the average, arrive at a target irradiated by this beam? Assume the beam to have uniform cross-section which is less than the target area),

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)

$\\n=\frac{P}{E}\\ n=\frac{9.42\times 10^{-3}}{3.14\times 10^{-19}}\\ n=3\times 10^{16}$

Mass of Hydrogen Atom (m)=1.67$\times$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

$\\v=\frac{p}{m}\\ v=\frac{1.05\times 10^{-27}}{1.67\times 10^{-27}}\\ v=0.628\ ms^{-1}$

Q: 11.5 The energy flux of sunlight reaching the surface of the earth is  $\dpi{100} 1.388\times 10^3 \hspace{1mm}W/m^2$. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of  $\dpi{100} 550 \hspace{1mm}nm$?

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

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{550\times 10^{-9}} \\E=3.61\times 10^{-19}\ J$

Energy flux(I) reaching the Earth's surface=1.388$\times$103 Wm-2

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

$\\n=\frac{I}{E}\\ n=\frac{1.388\times 10^{3}}{3.61\times 10^{-19}}\\ n=3.849\times 10^{21}\ m^{-2}$

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.

$h=slope\times e$

$h=4.12\times10^{-15}\times1.6\times10^{-19}$

$h=6.59210^{-34} Js$

The energy of a photon is given by

$\\E=\frac{hc}{\lambda } \\$

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

$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{589\times 10^{-9}}\\$

$E=3.37\times 10^{-19}\ J$

Power of the sodium lamp=100W

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

$\\R=\frac{P}{E}\\ R=\frac{100}{3.37\times 10^{-19}}\\ R=2.967\times 10^{20}\ s^{-1}$

Q: 11.8 The threshold frequency for a certain metal is  $\dpi{100} 3.3\times 10^1^4\hspace{1mm}Hz$  . If light of frequency   $\dpi{100} 8.2\times 10^1^4\hspace{1mm}Hz$  is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Threshold frequency of the given metal($\nu _{0}$)=  $\dpi{100} 3.3\times 10^1^4\hspace{1mm}Hz$

The work function of the given metal is

$\\\phi _{0}=h\nu _{0}\\ \phi _{0}=6.62\times 10^{-34}\times 3.3\times 10^{-14}\\ \phi _{0}=2.18\times 10^{-19}\ J$

The energy of the incident photons

$\\E=h\nu \\ E=6.62\times 10^{-34}\times 8.2\times 10^{14}\\ E=5.42\times 10^{-19}\ J$

Maximum Kinetic Energy of the ejected photoelectrons is

$\\E-\phi _{0}=3.24\times 10^{-19}\ J\\ E-\phi _{0}=2.025\ eV$

Therefore the cut off voltage is 2.025 eV

The energy of photons having 330 nm is

$\\E=\frac{hc}{\lambda } \\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{330\times 10^{-9}\times 1.6\times 10^{-19}}\\ E=3.7\ eV$

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

Q: 11.10  Light of frequency $\dpi{100} 7.21\times 10^1^4\hspace{1mm}Hz$  is incident on a metal surface. Electrons with a  maximum speed of  $\dpi{100} 6.0\times 10^5\hspace{1mm}m/s$ are ejected from the surface. What is the threshold  frequency for photoemission of electrons?

The energy of incident photons is E given by

$\\E=h\nu \\ E=6.62\times 10^{-34}\times 7.21\times 10^{14}\\ E=4.77\times 10^{-19}\ J$

Maximum Kinetic Energy of ejected electrons is

$\\KE_{max}=\frac{1}{2}mv^{2}\\ KE_{max}=\frac{9.1\times 10^{-31}\times (6\times 10^{5})^{2}}{2} \\KE_{max}=1.64\times 10^{-19}\ J$

Work Function of the given metal is

$\phi _{0}=E-KE_{max}=3.13\times 10^{-19}\ J$

The threshold frequency is therefore given by

$\\\nu _{0}=\frac{\phi _{0}}{h}\\ \nu _{0}=4.728\times 10^{14}\ Hz$

The energy of incident photons is given by

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{488\times 10^{-9}\times 1.6\times 10^{-19}}\\ E=2.54\ eV$

Cut-off potential is 0.38 eV

Work function is therefore, 2.54-0.38= 2.16 eV

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

$\\p=\sqrt{2mK}\\ p=\sqrt{2\times 9.1\times 10^{-31}\times 56\times 1.6\times 10^{-19}}\\ p=4.038\times 10^{-24}\ kg\ m\ s^{-1}$

De Broglie wavelength is given by the De Broglie relation as

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{4.038\times 10^{-24}}\\ \lambda =0.164\ nm$

the wavelength is 0.164 nm

Q: 11.13 (a) What is the momentum of an electron with kinetic energy of  $\dpi{100} 120\hspace{1mm}eV$.

The relation between momentum and kinetic energy is

$\\p=\sqrt{2mK}\\ p=\sqrt{2\times 9.1\times 10^{-31}\times 120\times 1.6\times 10^{-19}}\\ p=5.911\times 10^{-24}\ kg\ m\ s^{-1}$

Q: 11.13 (b) What is the speed of an electron with kinetic energy of $\dpi{100} 120\hspace{1mm}eV$.

The relation between speed and kinetic energy of a particle is

$\\v=\sqrt{\frac{2K}{m}}\\ v=\sqrt{\frac{2\times 120\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}\\ v=6.495\times 10^{6}\ m\ s^{-1}$

Q: 11.13 (c)  What is the de Broglie wavelength of an electron with kinetic energy of $\dpi{100} 120\hspace{1mm}eV$

De Broglie wavelength is given by

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{5.911\times 10^{-24}}\\ \lambda =1.12\times 10^{-10}\ m\\$

The de Broglie wavelength associated with the electron is 0.112 nm

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

$\\p=\frac{h}{\lambda }\\ p=\frac{6.62\times 10^{-34}}{589\times 10^{-9}}\\ p=1.12\times 10^{-27}\ kg\ m\ s^{-1}$

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

$\\K=\frac{p^{2}}{2m_{e}}\\ K=\frac{(1.12\times 10^{-27})^{2}}{2\times 9.1\times 10^{-31}}\\ K=6.89\times 10^{-25}\ J$

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

The kinetic energy of neutron would be

$\\K=\frac{p^{2}}{2m_{n}}\\ K=\frac{(1.12\times 10^{-27})^{2}}{2\times 1.675\times 10^{-27}}\\ K=3.74\times 10^{-28}\ J$

The momentum of the bullet is

$\\p=mv\\ p=0.04\times 10^{3}\\ p=40\ kg\ m\ s^{-1}$

De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{40}\\ \lambda =1.655\times 10^{-35}\ m$

The momentum of the ball is

$\\p=mv\\ p=0.06\ kg\ m\ s^{-1}$

De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{0.06}\\ \lambda =1.1\times 10^{-32}\ m$

The momentum of the dust particle is

$\\p=mv\\ p=10^{-9}\times 2.2\\ p=2.2\times 10^{-9} kg\ m\ s^{-1}$

De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{2.2\times 10^{-9} }\\ \lambda =3.01\times 10^{-25}\ m$

Q: 11.16 (a) An electron and a photon each have a wavelength of  $\dpi{100} 1.00\hspace{1mm}nm$ . Find their momenta.

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

$\\p=\frac{h}{\lambda }\\ p=\frac{6.62\times 10^{-34}}{10^{-9}}\\ p=6.62\times 10^{-25}kg\ m\ s^{-1}$

The energy of the photon is given by

$\\E=\frac{hc}{\lambda }\\$

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

$E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{10^{-9}}\\$

$E=1.86\times 10^{-16}\ J$

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

$\\K=\frac{p^{2}}{2m_{e}}\\ K=\frac{(6.62\times 10^{-25})^{2}}{2\times 9.1\times 10^{-31}}\\ K=2.41\times 10^{-19}\ J$

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

$\\p=\frac{h}{\lambda }\\ p=\frac{6.62\times 10^{-34}}{1.4\times 10^{-10}}\\ p=4.728\times 10^{-24}kg\ m\ s^{-1}$

The kinetic energy K would therefore be

$\\K=\frac{p^{2}}{2m}\\ K=\frac{(4.728\times 10^{-24})^{2}}{2\times 1.675\times 10^{-27}}\\ K=6.67\times 10^{-21}J$

The kinetic energy of the neutron is

$\\K=\frac{3}{2}kT\\ K=\frac{3}{2}\times 1.38\times 10^{-23}\times 300\\ K=6.21\times 10^{-21} J$

Where k Boltzmann's Constant is 1.38$\times$10-23 J/K

The momentum of the neutron will be p

$\\p=\sqrt{2m_{N}K}\\ p=\sqrt{2\times 1.675\times 10^{-27}\times 6.21\times 10^{-21}}\\ p=4.56\times 10^{-24}kg\ m\ s^{-1}$

Associated De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{4.56\times 10^{-24}}\\ \lambda =1.45\times 10^{-10} m$

De Broglie wavelength of the neutron is 0.145 nm.

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

E=pc

we also know

$E=h\nu$

Therefore the De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{h}{E/c}\\ \lambda =\frac{hc}{h\nu }\\ \lambda =\frac{c}{\nu }$

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

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

$\\K=\frac{3}{2}\times 1.38\times 10^{-23}\times 300\\ K=6.21\times 10^{-21}J$

Mass of Nitrogen molecule = 2$\times$14.0076$\times$1.66$\times$10-27=4.65$\times$10-26kg

The momentum of the molecule is

$\\p=\sqrt{2mK}\\ p=\sqrt{2\times 4.65\times 10^{-26}\times 6.21\times 10^{-21}}\\ p=2.4\times 10^{-23}kg\ m\ s^{-1}$

Associated De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{2.4\times 10^{-23}}\\ \lambda= 2.75\times 10^{-11}\ m$

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

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

$\\v=\sqrt{\frac{2K}{m_{e}}}\\ v=\sqrt{\frac{2eV}{m_{e}}}\\ v=\sqrt{2\times 1.76\times 10^{11}\times 500}\\ v=1.366\times 10^{7}ms^{-1}$        Specific charge is e/me=1.366$\times$1011C/kg

Using the same formula we get the speed of electrons to be 1.88$\times$109 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

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where m is the relativistic mass, m0 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 force due to the magnetic field on the electron will be Fb=evB (since the angle between the velocity and magnetic field is 90o)

This Facts as the centripetal force required for circular motion. Therefore

$\\F_{b}=\frac{mv^{2}}{r}\\ evB=\frac{mv^{2}}{r}\\ r=\frac{mv}{eB}\\ r=\frac{5.2\times 10^{6}}{1.76\times 10^{11}\times 1.3\times 10^{-4}}\\ r=0.227 m$

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

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where m is the relativistic mass, m0 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

$r=\frac{m_{e}v}{eB\sqrt{1-\frac{v^{2}}{c^{2}}}}$

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

$v=\sqrt{\frac{2eV}{m_{e}}}$

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 Fb=evB

This magnetic force acts as a centripetal force. Therefore

$\\\frac{m_{e}v^{2}}{r}=evB\\ \\\frac{m_{e}v}{r}=eB\\ \frac{m_{e}}{r}\times \sqrt{\frac{2eV}{m_{e}}}=eB\\ \sqrt{\frac{e}{m_{e}}}=\frac{\sqrt{2V}}{Br}\\ \frac{e}{m_{e}}=\frac{2V}{r^{2}B^{2}}\\ \frac{e}{m_{e}}=\frac{2\times 100}{(2.83\times 10^{-4})^{2}\times (0.12)^{2}}\\ \frac{e}{m_{e}}=1.73\times 10^{11}C\ kg^{-1}$

The wavelength of photons with maximum energy=0.45$A^{\circ}$

Energy of the photons is

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{0.45\times 10^{-10}}\\ E=4.413\times 10^{-15} J\\ E=27.6\ keV$

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.

Q: 11.24 In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy  $\dpi{100} 10.2\hspace{1mm}BeV$ into two $\dpi{100} \gamma$-rays of equal energy. What is the wavelength associated with each $\dpi{100} \gamma$-ray?  ($\dpi{100} 1\hspace{1mm}BeV=10^9\hspace{1mm}eV$)

The total energy of 2 $\dpi{100} \gamma$ rays=10.2 BeV

The average energy of 1 $\dpi{100} \gamma$ ray, E=5.1 BeV

The wavelength of the gamma-ray is given by

$\\\lambda =\frac{c}{\nu }\\\lambda =\frac{hc}{h\nu }\\ \lambda =\frac{hc}{E}\\ \lambda =\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{5.1\times 10^{9}\times 1.6\times 10^{-19}}\\ \lambda =2.436\times 10^{-16}m$

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

Wavelengths of photons being emiited=500 m

The energy of one photon is E

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{500}\\ E=3.96\times 10^{-28}J$

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

$\\n=\frac{P}{E}\\ n=\frac{10000}{3.96\times 10^{-28}}\\ n=2.525\times 10^{31}s^{-1}$

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

Area of the pupil(A)=0.4 cm2=4$\times$10-5 m2

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

P=IA

P=10-10$\times$4$\times$10-5

P=4$\times$10-15W

The average frequency of white light($\nu$)=6$\times$1014 Hz

The average energy of a photon in white light is

$\\E=h\nu \\ E=6.62\times 10^{-34}\times 6\times 10^{14}\\ E=3.972\times 10^{-19} J$

Number of photons reaching our eyes is n

$\\n=\frac{P}{E}\\ n=\frac{4\times 10^{-15}}{3.972\times 10^{-19}}\\ n=1.008\times 10^{4}s^{-1}$

Q: 11.26 Ultraviolet light of wavelength $\dpi{100} 2271\hspace{1mm}\dot{A}$ from a  $\dpi{100} 100\hspace{1mm}W$ mercury source irradiates a  photo-cell made of molybdenum metal. If the stopping potential is $\dpi{100} -1.3\hspace{1mm}V$, estimate the   work function of the metal. How would the photo-cell respond to a high intensity    $\dpi{100} (\sim 10^5\hspace{1mm}Wm^2)$  red light of wavelength $\dpi{100} 6382\hspace{1mm}\dot{A}$ produced by a $\dpi{100} He-Ne$ laser?

The energy of the incident photons is E given by

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{2271\times 10^{-10}\times 1.6\times 10^{-19}}\\ E=5.465\ eV$

Since stopping potential is -1.3 V work function is

$\\\phi _{0}=5.465-1.3\\ \phi _{0}=4.165 eV$

The energy of photons which red light consists of is ER

$\\E_{R}=\frac{hc}{\lambda _{R}}\\ E_{R}=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{6382\times 10^{-10}\times 1.6\times 10^{-19}}\\ E_{R}=1.945eV$

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.

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

$\\E_{1}=\frac{hc}{\lambda }\\ E_{1}=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{640.2\times 10^{-9}\times 1.6\times 10^{-19}}\\ E_{1}=1.939eV$

Stopping potential is 0.54 V

Work function is therefore

$\\\phi _{0}=1.939-0.54\\\phi _{0}=1.399 eV$

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

The energy of photons emitted by the ion source is

$\\E_{2}=\frac{hc}{\lambda }\\ E_{2}=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{427.2\times 10^{-9}\times 1.6\times 10^{-19}}\\ E_{2}=2.905eV$

New stopping voltage is

$\\E_{2}-\phi _{0}=2.905-1.399=1.506V$

Q: 11.28 A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

$\dpi{100} \lambda _1=3650\hspace{1mm}\dot{A},\lambda _2=4047\hspace{1mm}\dot{A}, \lambda _3=4358\hspace{1mm}\dot{A}, \lambda _4=5461\hspace{1mm}\dot{A},\lambda _5=6907\hspace{1mm}\dot{A}.$

The stopping voltages, respectively, were measured to be

$\dpi{100} V _0_1=1.28\hspace{1mm}V,V _0_2=0.95\hspace{1mm}V,V _0_3=0.74\hspace{1mm}V, V_0_4=0.16\hspace{1mm}V,V _0_5=0\hspace{1mm}V.$

Determine the value of Planck’s constant  $h$, the threshold frequency and work function for the material.

$\\h\nu =\phi _{0}+eV\\ V=(\frac{h}{e})\nu -\phi _{0}\\$

where V is stopping potential, h is planks constant, e is electronic charge, $\nu$ is frequency of incident photons and $\phi _{0}$ 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

$\nu_{1}=\frac{c}{\lambda_{1} }=\frac{3\times 10^{8}}{3650\times 10^{-10}}=8.219\times 10^{14}\ Hz$

$\nu_{2}=\frac{c}{\lambda_{2} }=\frac{3\times 10^{8}}{4047\times 10^{-10}}=7.412\times 10^{14}\ Hz$

$\nu_{3}=\frac{c}{\lambda_{3} }=\frac{3\times 10^{8}}{4358\times 10^{-10}}=6.884\times 10^{14}\ Hz$

$\nu_{4}=\frac{c}{\lambda_{4} }=\frac{3\times 10^{8}}{5461\times 10^{-10}}=5.493\times 10^{14}\ Hz$

$\nu_{5}=\frac{c}{\lambda_{5} }=\frac{3\times 10^{8}}{6907\times 10^{-10}}=4.343\times 10^{14}\ Hz$

$\dpi{100} V _0_1=1.28\hspace{1mm}V,V _0_2=0.95\hspace{1mm}V,V _0_3=0.74\hspace{1mm}V, V_0_4=0.16\hspace{1mm}V,V _0_5=0\hspace{1mm}V.$

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

$\\h=\frac{\left ( 1.28-0.16 \right )\times 1.6\times 10^{-19}}{(8.214-5.493)\times 10^{14}}\\ h=6.573\times 10^{-34}\ Js$

Planks constant calculated from the above chart is therefore $6.573\times 10^{-34}\ Js$

Which of these metals will not give photoelectric emission for a radiation of wavelength $\dpi{100} 3300\hspace{1mm}\dot{A}$ from a $\dpi{100} He-Cd$   laser placed $\dpi{100} 1\hspace{1mm}m$ away from the photocell? What happens if the laser is brought nearer and placed $\dpi{100} 50\hspace{1mm}cm$ away?

The wavelength of the incident photons=$3300\dot{A}$

The energy of the incident photons is

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{3300\times 10^{-10}\times 1.6\times 10^{-19}}\\ E=3.16 eV$

Mo and Ni will not give photoelectric emission for radiation of wavelength $\dpi{100} 3300\hspace{1mm}\dot{A}$ from a $\dpi{100} He-Cd$.

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.

Intensity of Incident light(I) =$\dpi{100} 10^-^5\hspace{1mm}Wm^-^2$

The surface area of the sodium photocell (A)=2 cm2= 2$\times$10-4 m2

The rate at which energy falls on the photo cell=IA=2$\times$10-9 W

The rate at which each of the 5 surfaces absorbs energy= IA/5=4$\times$10-10W

Effective atomic area of a sodium atom (A')= 10-20 m2

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

$\\R=\frac{IA}{5}\times \frac{A'}{A}\\ R=\frac{10^{-5}\times 10^{-20}}{5}\\ R=2\times 10^{-26}J/s$

The time required for photoelectric emission is

$\\t=\frac{\phi _{0}}{R}\\ t=\frac{2\times 1.6\times 10^{-19}}{2\times 10^{-26}}\\ t=1.6\times 10^{7}s\\ t\approx 0.507 \ years$

According to De Broglie's equation

$p=\frac{h}{\lambda }$

The kinetic energy of an electron with De Broglie wavelength $\dpi{100} 1\hspace{1mm}\dot{A}$ is given by

$\\K=\frac{p^{2}}{2m_{e}}\\ K=\frac{h^{2}}{\lambda ^{2}2m_{e}} \\K=\frac{(6.62\times 10^{-34})^{2}}{2\times 10^{-20}\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}}\\ K=149.375\ eV$

The kinetic energy of photon having wavelength $\dpi{100} 1\hspace{1mm}\dot{A}$ is

$\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{10^{-10}\times 1.6\times 10^{-19}}\\ E=12.375keV$

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

Kinetic energy of the neutron(K)=150eV

De Broglie wavelength associated with the neutron is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{h}{\sqrt{2m_{N}K }}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{2\times 1.675\times 10^{-27}\times 150\times 1.6\times 10^{-19}}}\\ \lambda =2.327\times 10^{-12}m$

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.

Absolute temperature = 273+27=300K

Boltzmann's Constant=1.38$\times$10-23 J/mol/K

The de Broglie wavelength associated with the neutron is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{h}{\sqrt{2m_{N}K}} \\\lambda =\frac{h}{\sqrt{3kT}}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{3\times 1.38\times 10^{-23}\times 300}}\\ \lambda =1.446 \dot{A}$

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.

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

$\\\lambda =\frac{h}{\sqrt{2m_{e}K}}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}\times 50000}}\\ \lambda =5.467\times 10^{-12} m$

The wavelength of yellow light = 5.9$\times$10-7 m

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

(Rest mass energy of electron $\dpi{100} =0.511\hspace{1mm}MeV$.)

Rest mass of the electron

$=mc^2=0.511MeV$

Momentum

$P=\frac{h}{\lambda}$

$=\frac{6.63\times ^{-34}}{10^{-15}}$

using the relativistic formula for energy

$E^2=(CP)^2+(mc^2)^2$

$=(3\times10^8 \times 6.63\times 10^{-19})^2+(0.511\times1.6\times10^{-19})^2$

$\approx 1.98\times10^{-10} J$

The kinetic energy K of a He atom is given by

$K=\frac{3}{2}kT$

mHe i.e. mass of one atom of He can be calculated as follows

$\\m_{He}=\frac{4\times 10^{-3}}{N_{A}}\\ =\frac{4\times 10^{-3}}{6.023\times 10^{23}}\\ m_{He}=6.64\times 10^{-27}\ kg$                             (NA is the Avogadro's Number)

De Broglie wavelength is given by

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{h}{\sqrt{2m_{He}K}}\\ \lambda =\frac{h}{\sqrt{3m_{He}kT}}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{3\times 6.64\times 10^{-27}\times 1.38\times 10^{-23}\times 300}}\\ \lambda =7.27\times 10^{-11}\ m$

The mean separation between two atoms is given by the relation

$\\d=\left ( \frac{V}{N} \right )^{\frac{1}{3}}\\$

From the ideal gas equation we have

$\\PV=nRT\\ PV=\frac{NRT}{N_{A}}\\ \frac{V}{N}=\frac{RT}{PN_{A}}$

The mean separation is therefore

$\\d=\left ( \frac{RT}{PN_{A}} \right )^{\frac{1}{3}}\\ d=\left ( \frac{kT}{P} \right )^{\frac{1}{3}}\\ d=\left ( \frac{1.38\times 10^{-23}\times 300}{1.01\times 10^{5}} \right )^{\frac{1}{3}}\\ d=3.35\times 10^{-9}\ m$

The mean separation is greater than the de Broglie wavelength.

The de Broglie wavelength associated with the electrons is

$\\\lambda =\frac{h}{\sqrt{3m_{e}kT}}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{3\times 9.1\times 10^{-31}\times 1.38\times 10^{-23}\times 300}}\\ \lambda =6.2\times 10^{-9}\ nm$

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

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

The speed of a charged particle is given by the relations

$v=\sqrt{2K\left ( \frac{e}{m} \right )}$

or

$v=Br\left ( \frac{e}{m} \right )$

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

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.

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.

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 $\nu\lambda$and .$\nu$ to have no physical significance as such.

The group speed is defined as

$V_{G}=\frac{h}{\lambda m}$

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 for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions NCERT solutions for class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei Solutions of NCERT class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

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.