When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \mathrm{~V}$ potential is…
When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \mathrm{~V}$ potential is required to stop the emitted electrons. Then, the work-function of the material is nearly
( given charge on electron $=1.602 \times 10^{-19} \mathrm{C}$ and Planck's constant $=6.625 \times 10^{-34} \mathrm{Js}$ )
$1.1 \mathrm{eV}$
$2.2 \mathrm{eV}$
$0.53 \mathrm{eV}$
$1 \mathrm{eV}$
Solution
Given, wavelength of light, $\lambda=4900 Å$
Stopping potential, $V=2 \mathrm{~V}$
Energy of the incident light,
$
\begin{aligned}
& E=\frac{12400 \mathrm{eV}}{\lambda Å} \\
\therefore \quad & E=\frac{12400}{4900} \mathrm{eV}=2.53 \mathrm{eV}
\end{aligned}
$
Kinetic energy, $\mathrm{KE}=$ Stopping potential $\times$ Charge of electron
$
=2 \times e=2 \mathrm{eV}
$
Using Einstein's photoelectric equation, Kinetic energy = Energy of incident light - Work function
$
\begin{aligned}
& 2=2.53-\phi \\
\Rightarrow \quad & \phi=2.53-2=0.53 \mathrm{eV}
\end{aligned}
$