Q155 / 160 · AP EAMCET 2023 (16 May Shift 2) · Physics
The threshold frequency of a photosensitive material is $\mathrm{v}$. When photons of frequency $2 \mathrm{v}$ incident on the material, photoelectrons are emitted with a maximum linear momentum P. To get photoelectrons with maximum linear momentum $2 \mathrm{P}$, the frequency of the incident photons is
- A$2 \mathrm{v}$
- B$3 \mathrm{v}$
- C$4 \mathrm{v}$
- Correct answerD$5 \mathrm{v}$
Solution
From the photoelectric equation
$\begin{aligned}
& h \times 2 v=h v+\frac{P^2}{2 m} \\
& h v=\frac{P^2}{2 m} ... (1)\\
& h v^{\prime}=h v+\frac{(2 p)^2}{2 m}\\
& h v^{\prime}=h v+\frac{4 p^2}{2 m} ... (2)
\end{aligned}$
Using the equation (1), we have
$\begin{aligned}
& h v^{\prime}=\frac{p^2}{2 m}+\frac{4 p^2}{2 m}=\frac{5 p^2}{2 m} \\
& h v^{\prime}=5 h v \\
& v^{\prime}=5 v
\end{aligned}$
Hence, the correct answer is option: 'D'.