### JEE Mains Previous Years Questions with Solutions

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### 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

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

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

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

### 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 $\left. {\,\,} \right]$

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