JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 10)
A photoelectric surface is illuminated successively by monochromatic light of wavelengths $$\lambda $$ and $${\lambda \over 2}.$$ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :
$${{hc} \over {3\lambda }}$$
$${{hc} \over {2\lambda }}$$
$${{hc} \over {\lambda }}$$
$${3\,{hc} \over {\lambda }}$$
Explanation
We know,
Einstein's photo electric equation,
(KE)max = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$0
In first case,
K = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$0 . . .(1)
In second case,
3K = $${{2hc} \over \lambda }$$ $$-$$ $$\phi $$0 . . .(2)
$$ \Rightarrow $$ $$3\left( {{{hc} \over \lambda } - {\phi _0}} \right)$$ = $${{2hc} \over \lambda } - {\phi _0}$$
$$ \Rightarrow $$ $${{3hc} \over \lambda }$$ $$-$$ 3$$\phi $$0 = $${{2hc} \over \lambda }$$ $$-$$ $$\phi $$0
$$ \Rightarrow $$ $${{hc} \over \lambda }$$ = 2$$\phi $$0
$$ \Rightarrow $$ $$\phi $$ = $${{hc} \over {2\lambda }}$$
Einstein's photo electric equation,
(KE)max = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$0
In first case,
K = $${{hc} \over \lambda }$$ $$-$$ $$\phi $$0 . . .(1)
In second case,
3K = $${{2hc} \over \lambda }$$ $$-$$ $$\phi $$0 . . .(2)
$$ \Rightarrow $$ $$3\left( {{{hc} \over \lambda } - {\phi _0}} \right)$$ = $${{2hc} \over \lambda } - {\phi _0}$$
$$ \Rightarrow $$ $${{3hc} \over \lambda }$$ $$-$$ 3$$\phi $$0 = $${{2hc} \over \lambda }$$ $$-$$ $$\phi $$0
$$ \Rightarrow $$ $${{hc} \over \lambda }$$ = 2$$\phi $$0
$$ \Rightarrow $$ $$\phi $$ = $${{hc} \over {2\lambda }}$$
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