Question:
In a photoelectric effect experiment the threshold wavelength of light is $380 \mathrm{~nm}$. If the wavelength of incident light is $260 \mathrm{~nm}$, the maximum kinetic energy of emitted electrons will be:
Given $\mathrm{E}($ in $\mathrm{eV})=\frac{1237}{\lambda(\text { in } \mathrm{nm})}$
Correct Option: 1
Solution:
(1) $\mathrm{KE}_{\max }=\mathrm{E}-\phi_{0}$
(where $\mathrm{E}=$ energy of incident light $\phi_{0}=$ work function)
$=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}$
$=1237\left[\frac{1}{260}-\frac{1}{380}\right]$
$=\frac{1237 \times 120}{380 \times 260}=1.5 \mathrm{eV}$
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