### JEE Mains Previous Years Questions with Solutions

4.5
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1

### AIEEE 2006

The time taken by a photoelectron to come out after the photon strikes is approximately
A
${10^{ - 4}}\,s$
B
${10^{ - 10}}\,s$
C
${10^{ - 16}}\,s$
D
${10^{ - 1}}\,s$

## Explanation

Emission of photo-electron starts from the surface after incidence of photons in about ${10^{ - 10}}s.$
2

### AIEEE 2006

The threshold frequency for a metallic surface corresponds to an energy of $6.2$ $eV$ and the stopping potential for a radiation incident on this surface is $5V.$ The incident radiation lies in
A
ultra-violet region
B
infra-red region
C
visible region
D
$x$-ray region

## Explanation

$\phi = 6.2\,eV = 6.2 \times 1.6 \times {10^{ - 19}}J$

$V = 5\,\,volt,\,\,{{hc} \over \lambda } - \phi = e{V_0}$

$\Rightarrow \lambda = {{hc} \over {\phi + e{V_0}}}$

$= {{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1.6 \times {{10}^{ - 19}}\left( {6.2 + 5} \right)}} \approx {10^{ - 7}}\,m$

This range lies in ultra violet range.
3

### AIEEE 2005

A photocell is illuminated by a small bright source placed $1$ $m$ away. When the same source of light is placed ${1 \over 2}$ $m$ away, the number of electrons emitted by photo-cathode would
A
increases by a factor of $4$
B
decreases by a factor of $4$
C
increases by a factor of $2$
D
decreases by a factor of $2$

## Explanation

$I \propto {1 \over {{r^2}}};{{{I_1}} \over {{I_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^2} = {1 \over 4}$

${I_2} \to 4\,\,$ times ${I_1}$

When intensity becomes 4 times, no. of photoelectrons emitted would increase by $4$ times, since number of electrons emitted per second is directly proportional to intensity.
4

### AIEEE 2005

If the kinetic energy of a free electron doubles, it's deBroglie wavelength changes by the factor
A
$2$
B
${1 \over 2}$
C
${\sqrt 2 }$
D
${1 \over {\sqrt 2 }}$

## Explanation

de-Broglie wavelength,

$\lambda = {h \over p} = {h \over {\sqrt {2.m,\left( {K.E} \right)} }}$

$\therefore$ $\lambda \propto {1 \over {\sqrt {K.E} }}$

If $K.E$ is doubled, wavelength becomes ${\lambda \over {\sqrt 2 }}$