1

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

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980$\mathop A\limits^ \circ$. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV$\mathop A\limits^ \circ$)
A
4a0
B
9a0
C
25a0
D
16a0

## Explanation

Energy of photon $= {{12500} \over {980}} = 12.75eV$

$\therefore$  Electron will excite to n = 4

Since 'R' $\propto$ n2

$\therefore$  Radius of atom will be 16a0
2

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

$\lambda ' = {{20} \over {27}}\lambda$
3

### 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
4

### 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$\beta$

NEET