NCERT

The vectors like velocity, displacement, and acceleration act along different directions in the two-dimensional motion. Resolving the vectors into the vertical and horizontal components allows the application of one-dimensional kinematic equations in each direction separately. It helps solve the pro

Explanation- velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ ^-1

Explanation-number of photon emitted per second n= $\frac{p}{\frac{hc}{?}}=\frac{p?}{hc}=\frac{20\times 5000\times {10}^{-10}}{6.62\times {10}^{-34}\times 3\times {10}^8}$

$=5\times {10}^{19}{s}^{-1}$

(ii) E=hc $/?$ = $\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{5000\times {10}^{-10}\times 1.6\times {10}^{-19}}=2.48eV$  this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4?d^2}\times ?r^2?t$ = ${?}_o$

$?t$ = $\frac{4?d^2}{p{r}^2}$ = $\frac{4\times 2\times 16\times 1.6\times {10}^{-19}\times 2^2}{20\times {(1.5\times {10}^{-10})}^{-2}}=28.4s$

(iv) N= $\frac{n?r^2}{4?d^2}\times ?t$

 = $\frac{5\times {10}^{19}\times {(1.5\times {10}^{-10})}^2\times 28.4}{4\times ({2)}^2}$ =2

(v) as the time of emission is 11.04s so

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_A{-m}_B}{m_A{+m}_B}$ v  and v₂= $\frac{2m_A}{m_A{+m}_B}$ v

_$?$ 

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

V_OB= 10 $\times$ 10¹⁴hz

$?=h{v}_0$

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

${?}_{OA}$ < ${?}_{OB}$

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

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

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

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

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}\times \mathrm{d}$ 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= $\frac{23}{0.97\times 6\times {10}^{26}}=3.95\times {10}^{-26}{\mathrm{m}}^3$

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

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

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

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

I = 100 $\times {10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}\times 3.3\times {10}^{16}\times 2.53\times {10}^{18}$ ( ${10}^{-4}$&nb

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.