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Q. In a photoelectric experiment ultraviolet light of wavelength $280 \,nm$ is used with lithium cathode having work function $\phi=2.5\, eV$. If the wavelength of incident light is switched to $400\, nm$, find out the change in the stopping potential.
$\left( h =6.63 \times 10^{-34} J\right.$, $c =3 \times 10^{8} \,ms ^{-1}$ )

JEE MainJEE Main 2021Dual Nature of Radiation and Matter

Solution:

$KE _{\max }= eV _{ S }=\frac{ hc }{\lambda}-\phi$
$\Rightarrow eV _{ s }=\frac{1240}{280}-2.5=1.93\, eV$
$\rightarrow V _{ S _{1}}=1.93\, V$... (i)
$\rightarrow eV _{ S _{2}}=\frac{1240}{400}-2.5=0.6\, eV$
$\Rightarrow V _{ S _{2}}=0.6 \, V \ldots$ (ii)
$\Delta V = V _{ S _{1}}- V _{ S _{2}}$
$=1.93-0.6=1.33\, V$