### JEE Mains Previous Years Questions with Solutions

4.5
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1

### 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.
2

### 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 ${\lambda \over {\sqrt 2 }}$
3

### AIEEE 2004

According to Einstein's photoelectric equation, the plot of the kinetic energy of the emitted photo electrons from a metal $Vs$ the frequency, of the incident radiation gives as straight the whose slope
A
depends both on the intensity of the radiation and the metal used
B
depends on the intensity of the radiation
C
depends on the nature of the metal used
D
is the same for the all metals and independent of the intensity of the radiation

## Explanation

From Equation $K.E = hv - \phi$

slope of graph of $K.E$ and $v$ is = $h$ (Plank's constant) which is same for all metals.
4

### AIEEE 2004

The work function of a substance is $4.0$ $eV.$ The longest wavelength of light that can cause photo-electron emission from this substance is approximately.
A
$310$ $nm$
B
$400$ $nm$
C
$540$ $nm$
D
$220$ $nm$

## Explanation

For the longest wavelength to emit photo electron

${{hc} \over \lambda } = \phi \Rightarrow \lambda = {{hc} \over \phi }$

$\Rightarrow \lambda = {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {40 \times 1.6 \times {{10}^{ - 16}}}} = 310nm$