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

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly
3.7 eV
3.2 eV
2.8 eV
2.5 eV

Selitys

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

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