1

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

If the de Broglie wavelength of an electron is equal to the 10–3 times the wavelength of a photon of frequency 6 $\times$ 1014 Hz, then the speed of electron is equal to : (Speed of light = 3 $\times$ 108 m/s, Planck's constant = 6.63 $\times$ 10–34 J.s, Mass of electron = 9.1 $\times$ 10–31 kg)
A
1.7 $\times$ 106 m/s
B
1.45 $\times$ 106 m/s
C
1.1 $\times$ 106 m/s
D
1.8 $\times$ 106 m/s

## Explanation

${h \over {mv}} = {10^{ - 3}}\left( {{{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}}} \right)$

v $= {{6.63 \times {{10}^{ - 34}} \times 6 \times {{10}^{14}}} \over {9.1 \times {{10}^{ - 31}} \times 3 \times {{10}^5}}}$

v $= 1.45 \times {10^6}$ m/s
2

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

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: (${{{hc} \over e}}$ = 1240 nm$-$V)
A
0.5 V
B
1.0 V
C
2.0 V
D
1.5 V

## Explanation

${{hc} \over {{\lambda _1}}} = \phi + e$V1        . . . (i)

${{hc} \over {{\lambda _2}}} = \phi + e$V2        . . . (ii)

(i) $-$ (ii)

hc$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$ = e(V1 $-$ V2)

$\Rightarrow$  V1 $-$ V2 = ${{hc} \over e}$$\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} \right)$

= (1240nm $-$ V) ${{100nm} \over {300nm \times 400nm}}$

= 1V
3

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

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths ${{{\lambda _A}} \over {{\lambda _B}}}$ is close to :
A
4.47
B
10.00
C
14.14
D
0.07

## Explanation

K.E. acquired by charge = K = qV

$\lambda$ = ${h \over P}$ = ${h \over {\sqrt {2mK} }}$ = ${h \over {\sqrt {2mqV} }}$

$\therefore$  ${{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50}$

$= 2 \times 7.07 = 14.14$
4

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

In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to :
A
2020 nm
B
250 nm
C
1700 nm
D
220 nm

## Explanation

The minimum wavelength of emitted photons is

$\lambda$ = ${{1240} \over {5.6 - 0.7}}nm$ = 250 nm