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1

### JEE Main 2021 (Online) 27th August Evening Shift

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?
A
0.96 V
B
1.25 V
C
0.24 V
D
1.5 V

## Explanation

$$k{E_{\max }} = {{hc} \over {{\lambda _i}}} + \phi$$

or $$e{V_o} = {{hc} \over {{\lambda _i}}} + \phi$$

when $$\lambda$$i = 670.5 nm ; Vo = 0.48

when $$\lambda$$i = 474.6 nm ; Vo = ?

So,

$$e(0.48) = {{1240} \over {670.5}} + \phi$$ ..... (1)

$$e({V_o}) = {{1240} \over {474.6}} + \phi$$ .....(2)

(2) $$-$$ (1)

$$e({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV$$

$${V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right)$$ Volts

Vo = 0.48 + 0.76

Vo = 1.24 V $$\simeq$$ 1.25 V
2

### JEE Main 2021 (Online) 27th August Morning Shift

In a photoelectric experiment, increasing the intensity of incident light :
A
increases the number of photons incident and also increases the K.E. of the ejected electrons
B
increases the frequency of photons incident and increases the K.E. of the ejected electrons
C
increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged
D
increases the number of photons incident and the K.E. of the ejected electrons remains unchanged

## Explanation

$$\to$$ Increasing intensity means number of incident photons are increased.

$$\to$$ Kinetic energy of ejected electrons depend on the frequency of incident photons, not the intensity.
3

### JEE Main 2021 (Online) 26th August Evening Shift

The de-Broglie wavelength of a particle having kinetic energy E is $$\lambda$$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?
A
$${1 \over 9}$$E
B
$${7 \over 9}$$E
C
E
D
$${16 \over 9}$$E

## Explanation

$$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$, $$mv = \sqrt {2mE}$$

$$\lambda \propto {1 \over {\sqrt E }}$$

$${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}$$, $${\lambda _2} = 0.75{\lambda _1}$$

$${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}$$

$${E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E$$ (E1 = E)

Extra energy given = $${{16} \over 9}E - E = {7 \over 9}E$$
4

### JEE Main 2021 (Online) 26th August Morning Shift

In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $$\phi$$ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 $$\times$$ 10$$-$$34 Js, c = 3 $$\times$$ 108 ms$$-$$1)
A
1.3 V
B
1.1 V
C
1.9 V
D
0.6 V

## Explanation

$$K{E_{\max }} = e{V_s} = {{hc} \over \lambda } - \phi$$

$$\Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 =$$ 1.93 eV

$$\Rightarrow {V_{{s_1}}} =$$ 1.93 V .... (i)

$$\Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 =$$ 0.6 eV

$$\Rightarrow {V_{{s_2}}} =$$ 0.6 V .... (ii)

$$\Delta$$V = $${V_{{s_1}}} - {V_{{s_2}}}$$ = 1.93 $$-$$ 0.6 = 1.33 V

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