Physics

Explanation -Given threshold frequency of A is given by v_0A= 5 $*{10}^{14}$ hz

V_OB= 10 $*$ 10¹⁴hz

$?=h{v}_0$

$\frac{{?}_{OA}}{{?}_{OB}}=\frac{5*{10}^{14}}{10*{10}^{14}}1$

${?}_{OA}$ < ${?}_{OB}$

$\left(ii\right)$  for metal A slope=h/e= $\frac{2}{(10-5){10}^{14}}$

$h=\frac{2e}{5*{10}^{14}}=\frac{2*1.6*{10}^{-19}}{5*{10}^{14}}$ = 6.4 $*{10}^{-34}$ js

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{B}$ slope=h/e= $\frac{2.5}{(15-10){10}^{14}}$ = 8 $*{10}^{-34}$ js

Explanation- F_x= $\frac{1}{4}\frac{q^2}{4?{?}_0{x}^2}$

W= ${?}_d^{?}fdx$ = ${?}_d^{?}\frac{q^{2}dx}{4*4?{?}_0}\frac{1}{x^2}$

= $\frac{q^2}{4*4?{?}_0}\frac{1}{d}$

= $\frac{{(1.6*{10}^{-19})}^2*9*{10}^9}{4*{10}^{-10}}$ J= 3.6eV 

Explanation- A= 10^-4m²

So d= 10^-3 and i= 100^-4A

I= 100W/m²

$\mathrm{?}=600\mathrm{n}\mathrm{m}=600*{10}^{-9}$

${\mathrm{?}}_{\mathrm{N}\mathrm{a}}=0.97\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}*\mathrm{d}$ 10^-4(10^-3)=10^-7m³

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}23\mathrm{k}\mathrm{g}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{m}$

So volume Na atoms=23/0.97m³

Volume occupied by one Na atom= $\frac{23}{0.97*6*{10}^{26}}=3.95*{10}^{-26}{\mathrm{m}}^3$

Number of Na atoms in target $\frac{{10}^{-7}}{3.95*{10}^{-26}}=2.53*{10}^{18}$

So energy falling per sec= $\frac{\mathrm{n}\mathrm{h}\mathrm{c}}{\mathrm{?}}=\mathrm{I}\mathrm{A}$

So n= $\frac{\mathrm{I}\mathrm{A}\mathrm{?}}{\mathrm{h}\mathrm{c}}$ = $\frac{100*{10}^{-4}*660*{10}^{-9}}{6.62*{10}^{-34}*3*{10}^8}=3.3*{10}^{16}$

N=P $*\mathrm{n}*\mathrm{N}\mathrm{a}=\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$

I = 100 $*{10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$ ( ${10}^{-4}$ A)

P= 7.48 $*{10}^{-21}$ it is less than 1.

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.

The NCERT Exemplar given on Shiksha's website provides a variety of numerical, conceptual and application-based questions around energy transitions, atomic models, and spectral series. It is a powerful study material for CBSE Board examination preparation and competitive exams like NEET and JEE.

According to Bohr's postulates, in specific stable orbits, electrons revolve without emitting energy. Energy is only absorbed or emitted when electrons jump between these orbits, preventing them from spiraling into the nucleus.