Q146 / 200 · NEET 2008 (Screening) · Physics
The work function of a surface of a photosensitive material is $6.2 \mathrm{eV}$. The wavelength of the incident radiation for which the stopping potential is $5 \mathrm{~V}$ lies in the
- Correct answerAultraviolet region
- Bvisible region
- Cinfrared region
- DX-ray region
Solution
According to laws of photoelectric effect
$\mathrm{KE}_{\max }=E-\phi$
where $\phi$ is work function and $\mathrm{KE}_{\max }$ is maximum kinetic energy of photoelectron.
$\begin{array}{l}
\therefore h v=e V_0+\phi \\
\text {or } h v=5 \mathrm{eV}+6.2 \mathrm{eV}=11.2 \mathrm{eV} \\
\therefore \lambda=\left(\frac{12400}{11.2}\right) Å \approx 1000 Å
\end{array}$
Hence, the radiation lies in ultraviolet region.
Hence, the correct answer is option: 'A'.