1

### JEE Main 2019 (Online) 11th January Evening Slot

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
A
${{25} \over {16}}$ $\lambda$
B
${{27} \over {20}}$ $\lambda$
C
${{16} \over {25}}$ $\lambda$
D
${{20} \over {27}}$ $\lambda$

## Explanation

For M $\to$ L steel

${1 \over \lambda }$ = K $\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$

for N $\to$ L

${1 \over {\lambda '}}$ = K$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$

$\lambda ' = {{20} \over {27}}\lambda$
2

### JEE Main 2019 (Online) 12th January Morning Slot

A particle of mass m moves in a circular orbit in a central potential field U(r) = ${1 \over 2}$ kr2. If Bohr 's quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as :
A
rn $\propto$ $\sqrt n$, En $\propto$ n
B
rn $\propto$ $\sqrt n$, En $\propto$ ${1 \over n}$
C
rn $\propto$ n, En $\propto$ n
D
rn $\propto$ n2, En $\propto$ ${1 \over {{n^2}}}$

## Explanation

Force due to this field, F = $- {{\partial U} \over {\partial r}}$

F = $- {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)$ = -kr

For circular orbit, ${{m{v^2}} \over r}$ = -kr

$\Rightarrow$ v $\propto$ r ..... (1)

Fron Bohr’s quantization condition

mvr = ${{nh} \over {2\pi }}$ .....(2)

From (1) and (2),

${r_n}$ $\propto$ ${n^{{1 \over 2}}}$

Given, U(r) = ${1 \over 2}$kr2

$\Rightarrow$ En = $- {1 \over 2}U\left( r \right)$ = $- {1 \over 4}k{r^2}$

$\Rightarrow$ En $\propto$ n
3

### JEE Main 2019 (Online) 12th January Evening Slot

In a radioactive decay chain, the initial nucleus is ${}_{90}^{232}$Th. At the end there are 6 $\alpha$-particles and 4 $\beta$-particles which are emitted. If the end nucleus is ${}_Z^A$X, A and Z are given by :
A
A = 208; Z = 80
B
A = 208; Z = 82
C
A = 200; Z = 81
D
A = 202; Z = 80

## Explanation

${}_{90}^{232}$Th  $\buildrel \, \over \longrightarrow$  ${}_{78}^{208}$Y   +   ${}_2^4$He

${}_{78}^{208}$Y   $\buildrel \, \over \longrightarrow$  ${}_{82}^{208}$ X   +  4$\beta$
4

### JEE Main 2019 (Online) 8th April Morning Slot

Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms fall on He+ ions in n = 1 and n = 2 states. The possible transition of helium ions as they absorb energy from the radiation is :
A
n = 1 $\to$ n = 4
B
n = 2 $\to$ n = 5
C
n = 2 $\to$ n = 4
D
n = 2 $\to$ n = 3

## Explanation

Energy released for tension n = 2 to n = 1 of hydrogen atom

$E = 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

Z = 1, n1 = 1, n2 = 2

$E = 13.6 \times 1 \times \left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$

$E = 13.6 \times {3 \over 4}eV$ = 10.2 eV

For He+ ion z = 2

(A) n = 1 to n = 4
$E = 13.6 \times {2^2} \times \left( {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{15} \over 4}eV$

(B) n = 2 to n = 4
$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = 13.6 \times {{3} \over 4}eV$

(C) n = 2 to n = 5
$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{5^2}}}} \right) = 13.6 \times {{21} \over 25}eV$

(D) n = 2 to n = 3
$E = 13.6 \times {2^2} \times \left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = 13.6 \times {{5} \over 9}eV$

So, possible transition is n = 2 $\to$ n = 4