JEE

PEC Chandigarh provide admissions to its Engineering BTech courses through the Junior Entrance Examination (JEE) cutoffs. Students can note that for the BTech, Integrated BTech, and BArch courses, the JoSAA cutoff 2024 was recorded between 10291 and 51699. Round 4 cutoff for BTech CSE was 10778 in 2023, while it was 10607 in 2022. The least competitive course for the 2024 cutoffs was BTech in Materials and Metallurgical Engineering with the cutoff of 51699.

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= $\frac{p}{\frac{hc}{?}}=\frac{p?}{hc}=\frac{20*5000*{10}^{-10}}{6.62*{10}^{-34}*3*{10}^8}$

$=5*{10}^{19}{s}^{-1}$

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

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

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

(iv) N= $\frac{n?r^2}{4?d^2}*?t$

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

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.

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

_$?$ initial=h/m_Av

_$?$ final=h/m_Av₁= $\frac{h\left(m_A{+m}_B\right)}{m_{a\left(m_A{-m}_B\right)v}}$

$d?$ = $?$ _final- $?$ _initial= $\frac{h}{m_{A}v}\{\frac{m_A{+m}_B}{m_A{-m}_B}-1\}$

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.

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