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

Edited By Vishal kumar | Updated on Sep 12, 2023 12:10 PM IST | #CBSE Class 12th

NCERT Solutions for Class 12 Physics Chapter 11 – Free PDF Download

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: Welcome to the updated for Class 12. On this Careers360 page, you'll find comprehensive class 12 physics chapter 11 exercise solutions written by subject experts in straightforward English. These solutions are also available in PDF format, allowing students to access them offline at their convenience.

Particles and the wave nature of matter are discussed NCERT Class 12 Physics chapter 11. In the NCERT CBSE Class 12 Physics chapter 11 solutions, you will study questions related to both the particle and wave nature of light. Experimental study of the photoelectric effect is an important topic of NCERT for the CBSE board exam. Dual Nature of Matter and Radiation Class 12 discuss this topic.

Try to solve all the questions in NCERT Class 12 Physics book yourself. If you are getting any doubt while solving questions, you can refer to chapter 11 physics class 12 ncert solutions. Understand all the graphs in NCERT syllabus for Class 12 Physics chapter 11. This will help to solve the questions from Dual Nature of Matter and Radiation Class 12 Physics NCERT chapter and to build an interest in the chapter 'Dual Nature of Radiation and Matter' .

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

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

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

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.

KE max =1.5 eV

KE mac =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 .

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

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

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 $\times$ 10 3 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

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

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 photo electrons 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.

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

The relation between momentum and kinetic energy is

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

The relation between speed and kinetic energy of a particle is

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

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

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

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 -26 kg

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.

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

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/m e =1.366 $\times$ 10 11 C/kg

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

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

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 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

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

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, 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

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

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 F b =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.

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 cm 2 =4 $\times$ 10 -5 m 2

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 -15 W

The average frequency of white light( $\nu$ )=6 $\times$ 10 14 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

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

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 E R

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

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

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

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

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 $6.573\times 10^{-34}\ Js$

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 cm 2 = 2 $\times$ 10 -4 m 2

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 -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

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

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 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 $\dpi{100} =0.511\hspace{1mm}MeV$ .)

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

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

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

$\\m_{He}=\frac{4\times 10^{-3}}{N_{A}}\\ =\frac{4\times 10^{-3}}{6.023\times 10^{23}}\\ m_{He}=6.64\times 10^{-27}\ kg$ (N A 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.

Dual Nature of Matter and Radiation Class 12 Solutions

There are 37 questions discussed in Dual Nature of Radiation and Matter NCERT solutions. The exercise questions of ch 11 Physics Class 12 is divided into two parts: exercise and additional exercise. Questions number 20 to 37 of Dual Nature of Matter and Radiation Class 12 are part of additional exercises.

NCERT Solutions for Class 12 Physics Chapter-wise

Below, we have provided the exercise-wise NCERT textbook solutions and questions:

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism NCERT solutions for class 12 physics chapter 5 Magnetism and Matter NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current NCERT solutions for class 12 physics chapter8 Electromagnetic Waves 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 NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits
JEE Main Highest Scoring Chapters & Topics
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Class 12 Physics ch 11 NCERT Solutions: Important Formulas and Diagrams

• Energy of photon(E): E=hv=hc

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Where: v= frequency of photon and = wavelength of photon

• Momentum of photon: p= h

Where: h= plank’s constant and is wavelength of photon

• Velocity of charged particle passing through a potential v:

v= 2eVm

Where: e= charge of electron and m is the mass of electron

• Einstein’s Photoelectric Equation: Energy Quantum of Radiation

Where: Kmax is the maximum kinetic energy of the emitted electron and is the work function of metal

Dual Nature of Matter and Radiation Class 12-Topics

The topic discussed in the Class 12 NCERT Physics chapter Dual Nature of Matter and Radiation are Electron Emission, Photoelectric Effect and It's Experimental Study, Wave Theory of Light and Photoelectric Effect, Einsteins Photo Electric Equation, Particle Nature of Light, Wave Nature of Matter and Davisson and Germer Experiment.

Importance of NCERT Solutions for Class 12 Physics Chapter 11

• The chapter Dual Nature of Radiation and Matter comes under modern Physics part of Class 12 NCERT book. NCERT Class 12 Physics Solutions is important for competitive exams like NEET and JEE Mains.
• As far as CBSE 12th board exam is considered, 3 to 5 marks questions are expected from this chapter. In CBSE board exam 2019, 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.

Key Feathers of Class 12 Physics ch 11 NCERT Solutions

1. Comprehensive Coverage: These class 12 physics chapter 11 exercise solutions encompass all the topics and questions presented in Chapter 11, ensuring a thorough understanding of chapter 11.

2. Detailed Explanations: Each chapter 11 physics class 12 ncert solutions offers detailed step-by-step explanations, simplifying complex concepts for students.

3. Clarity and Simplicity: The dual nature of radiation and matter ncert solutions are presented in clear and simple language, ensuring ease of understanding.

4. Practice Questions: Exercise questions are included for practice and self-assessment.

5. Exam Preparation: These class 12 chapter 11 physics ncert solutions are essential for board exam preparation and provide valuable support for competitive exams.

6. Foundation for Advanced Study: Concepts covered in this chapter serve as the foundation for more advanced studies in Radiation and Matter physics and related fields.

7. Free Access: These class 12 physics chapter 11 ncert solutions are available for free, ensuring accessibility to all students.

Also, check NCERT Exemplar Class 12 Solutions

1. What is the weightage of chapter Dual Nature of Radiation and Matter for CBSE board exam?

Around 3 to 5 marks questions can be expected from the NCERT Class 12 chapter 11 for CBSE board exam according to the previous year papers. Students can also use the help of NCERT Exemplar Problems for Physics for more problems.

2. Which topics are important from NCERT Class 12 Physics chapter 11 solutions?

All the topics of the chapter are important. Understand all the graphs mentioned in the NCERT book. Also understand all the formulas listed in the chapter. The CBSE papers are based on the NCERT Syllabus. So all the concepts listed in the chapter are important.

3. What is wave-particle duality in dual nature of matter and radiation class 12?

Wave-particle duality refers to the ability of particles of matter, such as electrons and photons, to exhibit both wave-like and particle-like behaviour. This means that they can display properties of both waves, such as diffraction and interference, and particles, such as quantization and the photoelectric effect.

4. How dual nature of light class 12 is important for NEET?

dual nature ncert solutions physics is important for NEET as it helps students understand the behaviour of particles and light on a microscopic level, including the properties of atoms and molecules, and how light interacts with different materials. This understanding is crucial for understanding key topics in NEET physics such as the behaviour of electrons in solids and the behaviour of light in different materials.

5. What do you mean by photoelectric effect in class 12 physics chapter 11 ncert solutions?

The photoelectric effect is a phenomenon in which electrons are emitted from a metal surface when light shines on it. The effect occurs because the light is made up of individual packets of energy, called photons, which transfer energy to electrons in the metal surface, causing them to be emitted.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9