Q150 / 200 · NEET 2005 · Physics
The photosensitive metallic surface has work function $h v_0$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 \mathrm{~m} / \mathrm{s}$. When the photon energy is increased $5 h v_0$. Then maximum velocity of photoelectrons will be:
- A$2 \times 10^7 \mathrm{~m} / \mathrm{s}$
- B$2 \times 10^6 \mathrm{~m} / \mathrm{s}$
- Correct answerC$8 \times 10^6 \mathrm{~m} / \mathrm{s}$
- D$8 \times 10^5 \mathrm{~m} / \mathrm{s}$
Solution
According to the question
$$
\begin{gathered}
2 h v_0=h v_0+\frac{1}{2} m\left(4 \times 10^6\right)^2 \\
5 h v_0=h v_0+\frac{1}{2} m v^2 \\
\Rightarrow 4 \times \frac{1}{2} m\left(4 \times 10^6\right) \\
\Rightarrow \frac{1}{2} m v^2 \\
v=8 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{gathered}
$$
Hence, the correct answer is option: 'C'.