Question 237: Dual Nature of Matter and Radiation

Question

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 __________.

Accepted Answer

$${{hc} \over \lambda } - \phi $$ = eV .....(i)
$${{hc} \over {3\lambda }} - \phi  = {{eV} \over 4}$$ .....(ii)
From (i) and (ii),
$${{hc} \over {3\lambda }} - \phi  =$$ $${{hc} \over {4\lambda }} - {\phi  \over 4}$$
$$ \Rightarrow $$ $${{hc} \over \lambda }\left( {{1 \over 3} - {1 \over 4}} ight) = {{3\phi } \over 4}$$
$$ \Rightarrow $$ $${{hc} \over {9\lambda }} = \phi $$
Also, $$\phi $$ = $${{hc} \over {{\lambda _0}}}$$
$$ 	herefore $$ $${{hc} \over {9\lambda }} = {{hc} \over {{\lambda _0}}}$$
$$ \Rightarrow $$ $$\phi $$ = 9$$\lambda $$
So, n = 9

Dual Nature of Matter and Radiation

Integer / Numeric Answer Type