JEE Advance - Physics (2014 - Paper 2 Offline - No. 7)

From Einstein's photoelectric equation,

kinetic energy of photoelectrons,

$$K = {1 \over 2}m{v^2} = {{hc} \over \lambda } - \phi $$ ..... (i)

For $$\lambda$$ = 248 nm, v = u₁

$$\therefore$$ $${K_1} = {1 \over 2}mu_1^2 = {{hc} \over {248}} - \phi $$ ...... (ii)

For $$\lambda$$ = 310 nm, v = u₂

$$\therefore$$ $${K_2} = {1 \over 2}mu_2^2 = {{hc} \over {310}} - \phi $$ ....... (iii)

Divide eqn. (ii) by eqn. (iii),

$${{u_1^2} \over {u_2^2}} = {{{{hc} \over {248}} - \phi } \over {{{hc} \over {310}} - \phi }} = {{{{1240} \over {248}} - \phi } \over {{{1240} \over {310}} - \phi }}$$

$$ \Rightarrow {\left( {{2 \over 1}} \right)^2} = {{5 - \phi } \over {4 - \phi }}$$

$$ \Rightarrow 16 - 4\phi = 5 - \phi \Rightarrow 3\phi = 11$$

or, $$\phi$$ = 3.67 eV $$\approx$$ 3.7 eV