JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 21)
When radiation of wavelength $$\lambda $$ is used to
illuminate a metallic surface, the stopping
potential is V. When the same surface is
illuminated with radiation of wavelength 3$$\lambda $$,
the stopping potential is
$${V \over 4}$$. If the threshold
wavelength for the metallic surface is n$$\lambda $$ then
value of n will be __________.
Answer
9
Explanation
$${{hc} \over \lambda } - \phi $$ = eV .....(i)
$${{hc} \over {3\lambda }} - \phi = {{eV} \over 4}$$ .....(ii)
From (i) and (ii),
$${{hc} \over {3\lambda }} - \phi =$$ $${{hc} \over {4\lambda }} - {\phi \over 4}$$
$$ \Rightarrow $$ $${{hc} \over \lambda }\left( {{1 \over 3} - {1 \over 4}} \right) = {{3\phi } \over 4}$$
$$ \Rightarrow $$ $${{hc} \over {9\lambda }} = \phi $$
Also, $$\phi $$ = $${{hc} \over {{\lambda _0}}}$$
$$ \therefore $$ $${{hc} \over {9\lambda }} = {{hc} \over {{\lambda _0}}}$$
$$ \Rightarrow $$ $$\phi $$ = 9$$\lambda $$
So, n = 9
$${{hc} \over {3\lambda }} - \phi = {{eV} \over 4}$$ .....(ii)
From (i) and (ii),
$${{hc} \over {3\lambda }} - \phi =$$ $${{hc} \over {4\lambda }} - {\phi \over 4}$$
$$ \Rightarrow $$ $${{hc} \over \lambda }\left( {{1 \over 3} - {1 \over 4}} \right) = {{3\phi } \over 4}$$
$$ \Rightarrow $$ $${{hc} \over {9\lambda }} = \phi $$
Also, $$\phi $$ = $${{hc} \over {{\lambda _0}}}$$
$$ \therefore $$ $${{hc} \over {9\lambda }} = {{hc} \over {{\lambda _0}}}$$
$$ \Rightarrow $$ $$\phi $$ = 9$$\lambda $$
So, n = 9
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