1

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

The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 $\times$ 107)ct + sin(6.28 $\times$ 107)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(c = 3 $\times$ 108 ms$-$1, h = 6.6 $\times$ 10$-$34J-s)
A
6.82 eV
B
12.5 eV
C
8.52 eV
D
7.72 eV

## Explanation

Given that,

B = B0[sin (3.14 $\times$ 107) ct + sin(6.28 $\times$ 107) ct]

This light wave is non-monochromotic wave as here is two different frequency in the magnetic field.

Given work function ($\phi$) = 4.7 eV

Question says to find the maximum kinetic energy, so we have to use maximum frequency among the available frequency.

As,

Kmax = Emax $-$ $\phi$

Emax = hF

= h $\times$ ${\omega \over {2\pi }}$

= 6.6 $\times$ 10$-$34 $\times$ ${{6.28 \times {{10}^7} \times 3 \times {{10}^8}} \over {2\pi }}$

= 6.6$\times$ 3 $\times$ 10$-$19 J

= ${{6.6 \times 3 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}eV$

= 12.375 $eV$

$\therefore$  Kmax = 12.375 $-$ 4.7

= 7.675 $eV$

$\simeq$  7.7 $eV$
2

### JEE Main 2019 (Online) 10th January Evening Slot

A metal plate of area 1 $\times$ 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be
A
1014 and 10 eV
B
1012 and 5 eV
C
1011 and 5 eV
D
1010 and 5 eV

## Explanation

Given, $\phi$ = 5 eV, E = 10 eV

$\therefore$ KE = E – $\phi$ = 10 – 5 = 5 eV

Energy incident on the plate,

E' = $\left( {16 \times {{10}^{ - 3}}} \right)\left( {1 \times {{10}^{ - 4}}} \right)$

= 16 $\times$ 10-7 J/s

= 1013 eV/s

Number of photons emitted per second

= 10% of ${{E'} \over E} = 0.1 \times {{{{10}^{13}}} \over {10}}$ = 1011 s-1
3

### 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
4

### 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