Q139 / 180 · NEET 2026 (Re-NEET) · Physics
A ray of light with wavelength $\lambda$ is incident on three different photo-electric cells namely 1, 2 and 3. The threshold wavelength of these photo-electric cells are $\lambda_1$, $\lambda_2$, and $\lambda_3$, respectively and the magnitude of stopping potentials of these cells are $V_1$, $V_2$ and $V_3$, respectively. The relation between $\lambda$ and threshold wavelengths are $\lambda_1<\lambda$, $\lambda_2>\lambda$ and $\lambda_3\gg\lambda$. The correct option is :
- A$V_1 < V_2,\,V_3=0$
- Correct answerB$V_1=0,\,V_2 < V_3$
- C$V_1=0,\,V_2>V_3$
- D$V_1>V_2,\,V_3=0$
Solution
According to Einstein's photoelectric equation, the magnitude of the stopping potential $V$ is given by:
$eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_{th}}$
For photoelectric emission to take place, the incident wavelength $\lambda$ must be less than the threshold wavelength $\lambda_{th}$.
For cell 1, $\lambda > \lambda_1$. No photoelectric emission occurs, so $V_1 = 0$.
For cell 2, $\lambda < \lambda_2$. Photoelectric emission occurs, and $V_2 = \dfrac{hc}{e} \left( \dfrac{1}{\lambda} - \dfrac{1}{\lambda_2} \right) > 0$.
For cell 3, $\lambda \ll \lambda_3$. Photoelectric emission occurs, and $V_3 = \dfrac{hc}{e} \left( \dfrac{1}{\lambda} - \dfrac{1}{\lambda_3} \right) > 0$.
Since $\lambda_3 \gg \lambda$ and $\lambda_2 > \lambda$, it follows that $\lambda_3 > \lambda_2$, which gives $\dfrac{1}{\lambda_3} < \dfrac{1}{\lambda_2}$.
Therefore, $\dfrac{1}{\lambda} - \dfrac{1}{\lambda_3} > \dfrac{1}{\lambda} - \dfrac{1}{\lambda_2}$, which implies $V_3 > V_2$.
Thus, $V_1 = 0$ and $V_2 < V_3$.
Answer: $V_1=0,\,V_2 < V_3$
$eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_{th}}$
For photoelectric emission to take place, the incident wavelength $\lambda$ must be less than the threshold wavelength $\lambda_{th}$.
For cell 1, $\lambda > \lambda_1$. No photoelectric emission occurs, so $V_1 = 0$.
For cell 2, $\lambda < \lambda_2$. Photoelectric emission occurs, and $V_2 = \dfrac{hc}{e} \left( \dfrac{1}{\lambda} - \dfrac{1}{\lambda_2} \right) > 0$.
For cell 3, $\lambda \ll \lambda_3$. Photoelectric emission occurs, and $V_3 = \dfrac{hc}{e} \left( \dfrac{1}{\lambda} - \dfrac{1}{\lambda_3} \right) > 0$.
Since $\lambda_3 \gg \lambda$ and $\lambda_2 > \lambda$, it follows that $\lambda_3 > \lambda_2$, which gives $\dfrac{1}{\lambda_3} < \dfrac{1}{\lambda_2}$.
Therefore, $\dfrac{1}{\lambda} - \dfrac{1}{\lambda_3} > \dfrac{1}{\lambda} - \dfrac{1}{\lambda_2}$, which implies $V_3 > V_2$.
Thus, $V_1 = 0$ and $V_2 < V_3$.
Answer: $V_1=0,\,V_2 < V_3$
Hence, the correct answer is option: 'B'.