Q120 / 155 · WBJEE 2016 · Physics
The work function of Cesium is $2.27 \mathrm{eV}$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 \mathrm{nm}$ wavelength is
- A$0.5 \mathrm{V}$
- B$-0.2 \mathrm{V}$
- C$-0.5 \mathrm{V}$
- Correct answerDNone of the above
Solution
Work function of cesium $=2.27 \mathrm{eV}$
The wavelength of incident light
$$
\lambda=600 \mathrm{nm}=600 \times 10^{-9} \mathrm{m}
$$
The energy of incident photon $=h v$
$$
\begin{aligned}
E &=\frac{h c}{\lambda} \\
&=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9} \times 1.6 \times 10^{-19}} e \mathrm{V} \\
&=2.06 \mathrm{eV} \\
\frac{h c}{\lambda} < \phi
\end{aligned}
$$
$2.06 \mathrm{eV} < 2.27 \mathrm{eV}$
So, no emission of electrons will take place. Hence, none of the given options is correct.
Hence, the correct answer is option: 'D'.