Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Explanation- given d= 0.1nm,  $?$ =30⁰ and n=1

According to bragg's law

2dsin $?$ = n $?$

2 $*0.1*sin30=?$

$?=0.1nm={10}^{-10}$

$?=\frac{h}{mv}=\frac{h}{p}$

P=h/ $?$  = $\frac{6.62*{10}^{-34}}{{10}^{-10}}$ = 6.62 $*{10}^{-24}$ kgm/s

k.E= 1/2mv²= $\frac{1}{2}\frac{m^2{v}^2}{m}$ = $\frac{p^2}{2m}$ = 0.21eV

The de Broglie wavelength  depends on:
Planck's constant h (a universal constant),
velocity v of the particle
mass m of the particle.

Photons do not have enough energy ( )to overcome the work function (? ) of the material below the threshold frequency.The energy per photon remains too low to liberate electrons even if the light intensity is increased, so photoelectric emission cannot occur.

According to de Broglie's hypothesis, all matter exhibits wave-like behavior. It laid the foundation for quantum mechanics by introducing the concept of matter waves. The dual nature of particles (both wave and particle) is essential in designing technologies like electron microscopes and helps explain phenomena like electron diffraction.

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Explanation- p=100W

_{$\lambda$ 1}= 1nm

_{$\lambda$ 2}= 500nm

Also n₁and n₂ represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n₁hc/ $\lambda$ ₁₌n₂hc/ $\lambda$ ₂

So n₁/ $\lambda$ ₁ = n₂/ $\lambda$ ₂

So n₁/n₂= 1/500

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Explanation- In photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.Thus, all the electrons that absorb a photon doesn't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.

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Explanation- (i) Here it is given that, an electron absorbs 2 photons each of frequency ν then ν where, v′ is the frequency of emitted electron.

Given, E_max= hv- ${?}_0$

Now, maximum energy for emitted electrons is E_max= h2v- ${?}_0=2\mathrm{h}\mathrm{v}-{?}_0$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible

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Explanation- $\lambda =h/\sqrt[]{2mqv}$

$\lambda \propto \frac{1}{\sqrt[]{mq}}$

$\frac{{\lambda }_p}{{\lambda }_a}=\frac{\sqrt[]{m_a{q}_a}}{\sqrt[]{m_p{q}_p}}=\frac{\sqrt[]{4m_p*2e}}{\sqrt[]{m_p*e}}$ = $\sqrt[]{8}$

So wavelength of proton is $\sqrt[]{8}$ times of alpha particle