Q144 / 200 · NEET 2010 (Screening) · Physics
The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface, having work function $5.01 \mathrm{eV}$, when ultraviolet light of $200 \mathrm{~nm}$ falls on it, must be
- A$2.4 \mathrm{~V}$
- B$-1.2 \mathrm{~V}$
- C$-2.4 \mathrm{~V}$
- Correct answerD$1.2 \mathrm{~V}$
Solution
Energy of incident light $\mathrm{E}(\mathrm{eV})=\frac{12375}{2000}$ $=6.2 \mathrm{eV}(200 \mathrm{~nm}=2000 Å)$
According to the relation $\mathrm{E}=\mathrm{W}_0+\mathrm{eV}_0$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{V}_0 & =\frac{\mathrm{E}-\mathrm{W}_0}{\mathrm{e}} \\
& =\frac{(6.2-5.01) \mathrm{e}}{\mathrm{e}} \\
& =1.2 \mathrm{~V}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.