### JEE Mains Previous Years Questions with Solutions

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1

### 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.
2

### 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

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

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

### AIEEE 2003

Two identical photo-cathodes receive light of frequencies ${f_1}$ and ${f_2}$. If the velocities of the photo electrons (of mass $m$ ) coming out are respectively ${v_1}$ and ${v_2},$ then
A
$v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$
B
${v_1} + {v_2} = {\left[ {{{2h} \over m}\left( {{f_1} + {f_2}} \right)} \right]^{1/2}}$
C
$v_1^2 + v_2^2 = {{2h} \over m}\left( {{f_1} + {f_2}} \right)$
D
${v_1} - {v_2} = {\left[ {{{2h} \over m}\left( {{f_1} - {f_2}} \right)} \right]^{1/2}}$

## Explanation

For one photo cathode

$h{f_1} - W = {1 \over 2}mv_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

For another photo cathode

$h{f_2} - W = {1 \over 2}mv_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Subtracting $(ii)$ from $(i)$ we get

$\left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = {1 \over 2}mv_1^2 - {1 \over 2}mv_2^2$

$\therefore$ $h\left( {{f_1} - {f_2}} \right) = {m \over 2}\left( {v_1^2 - v_2^2} \right)$

$\therefore$ $v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$
4

### AIEEE 2002

Sodium and copper have work functions $2.3$ $eV$ and $4.5$ $eV$ respectively. Then the ratio of the wavelengths is nearest to
A
$1:2$
B
$4:1$
C
$2:1$
D
$1:4$

## Explanation

We know that work function is the energy required and energy $E = h\upsilon$

$\therefore$ ${{{E_{NA}}} \over {{E_{Cu}}}} = {{h{\upsilon _{Na}}} \over {h{\upsilon _{cu}}}} = {{{\lambda _{cu}}} \over {{\lambda _{Na}}}}$

$\left[ {\,\,} \right.$ as ${\,\,\,\upsilon \propto {1 \over \lambda }}$ for light $\left. {\,\,} \right]$

$\therefore$ ${{{\lambda _{Na}}} \over {{\lambda _{Cu}}}} = {{{E_{Cu}}} \over {{E_{Na}}}} = {{4.5} \over {2.3}} \approx {2 \over 1}$