1

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

The de-Broglie wavelength ($\lambda$B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($\lambda$G) by :
A
$\lambda$B = 2$\lambda$G
B
$\lambda$B = 3$\lambda$G
C
$\lambda$B = $\lambda$G/2
D
$\lambda$B = $\lambda$G/3
2

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

Surface of certain metal is first illuminated with light of wavelength $\lambda$1 = 350 nm and then, by light of wavelength $\lambda$2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :

(Energy of photon n = ${{1240} \over {\lambda (in\,mm)}}$eV)
A
1.8
B
2.5
C
5.6
D
1.4

## Explanation

Let speed of photon electron in first case is 2v

then in the second case speed is v.

For first case

${{hc} \over {{\lambda _1}}} = \phi + {1 \over 2}$m(2v)2

For second case,

${{hc} \over {{\lambda _2}}} = \phi + {1 \over 2}$mv2

$\therefore$   ${{{{hc} \over {{\lambda _1}}} - \phi } \over {{{hc} \over {{\lambda _2}}} - \phi }} = {{{1 \over 2}m \times 4{v^2}} \over {{1 \over 2}m{v^2}}}$

$\Rightarrow$   ${{{hc} \over {{\lambda _1}}} - \phi }$ = 4$\left( {{{hc} \over {{\lambda _2}}} - \phi } \right)$

$\Rightarrow$   ${{4hc} \over {{\lambda _2}}}$ $-$ ${{hc} \over {{\lambda _1}}}$ = 3$\phi$

$\Rightarrow$   $\phi$ $=$ ${{hc} \over 3}\left( {{4 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right)$

$\Rightarrow$   $\phi$ $=$ ${{1240} \over 3}\left( {{4 \over {540}} - {1 \over {350}}} \right)$

$\Rightarrow$   $\phi$ = 1.8 eV
3

### 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 $eV$
4

### 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