1

### JEE Main 2018 (Online) 15th April Evening Slot

The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :
A
0.529 $\mathop {\rm A}\limits^ \circ$
B
2$\pi$ $\times$ 0.529 $\mathop {\rm A}\limits^ \circ$
C
${{0.529} \over {2\pi }}$ $\mathop {\rm A}\limits^ \circ$
D
4 $\times$ 0.529 $\mathop {\rm A}\limits^ \circ$

## Explanation

Radius    r = 0.529 $\times$ ${{{n^2}} \over z}$

In first Bohr orbit of hydrogen atom radius,

r = 0.529 $\times$ ${{{1^2}} \over 1}$ = 0.529 $\mathop A\limits^ \circ$

Angular momentum,

mvr = ${{nh} \over {2\pi }}$

$\therefore\,\,\,$ for    n = 1,

mvr = ${h \over {2\pi }}$

$\Rightarrow$$\,\,\,$ 2$\pi$r = ${h \over {mv}}$ = $\lambda$   [ as   $\lambda$ = ${h \over {mv}}$]

$\therefore\,\,\,$ $\lambda$ = 2$\pi$r = 2 $\pi$ $\times$ 0.5 29
2

### JEE Main 2018 (Online) 16th April Morning Slot

Which of the following statements is false ?
A
Photon has momentum as well as wavelength.
B
Splitting of spectral lines in electrical field is called Stark effect.
C
Rydberg constant has unit of energy.
D
Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases.

## Explanation

When a black body is heated, black body emit high energy radiation, from higher wavelength to lower wavelength.
3

### JEE Main 2019 (Online) 9th January Morning Slot

The highest value of the calculated spin- only magnetic moment (in BM) among all the transition metal complexes is :
A
5.92
B
6.93
C
3.87
D
4.90

## Explanation

In transition metal contains d orbital, and in d orbital maximum no of unpaired electron possible = 5.

Spin only magnetic moment,

${\mu _{spin}} = \sqrt {n\left( {n + 2} \right)}$    B. M

here n $=$ Number of unpaired electrons.

$\therefore$    ${\mu _{spin}} = \sqrt {5\left( {5 + 2} \right)}$0    B. M

$=$ 5.92   B. M
4

### JEE Main 2019 (Online) 9th January Morning Slot

For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number $\left( {\overline v } \right)$ against $\left( {{1 \over {{n^2}}}} \right)$ will be (The Rydberg constant, RH is in wave number unit)
A
Linear with intercept $-$RH
B
Non linear
C
Linear with slope RH
D
Linear with slope $-$RH

## Explanation

We know,

$\overline v = {1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_f^2}} - {1 \over {n_i^2}}} \right]$

here ni $=$ 8 and nf $=$ n

Z $=$ 1 for hydrogen

$\therefore$   $\overline v = {1 \over \lambda } = {R_H}\left[ {{1 \over {{n^2}}} - {1 \over {{8^2}}}} \right]$

So, graph is $\to$ So, graph will be linear with slope RH.