Question:
When photons of wavelength $4000\text{ \AA}$ are incident on a photosensitive material of cut-off wavelength $4800\text{ \AA}$, the stopping potential is $V$. If the same photons are incident on another photosensitive material of cut-off wavelength $6000\text{ \AA}$, then the stopping potential is:
When photons of wavelength $4000\text{ \AA}$ are incident on a photosensitive material of cut-off wavelength $4800\text{ \AA}$, the stopping potential is $V$. If the same photons are incident on another photosensitive material of cut-off wavelength $6000\text{ \AA}$, then the stopping potential is:
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For photoelectric effect questions involving two different materials but the same incident radiation, write
\[
eV_s
=
hc
\left(
\frac{1}{\lambda}
-
\frac{1}{\lambda_0}
\right)
\]
for each material and compare the equations directly. The constants $h$, $c$, and $e$ cancel, making the calculation much simpler.
Updated On: Jun 15, 2026
- $1.5V$
- $0.5V$
- $4V$
- $2V$
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The Correct Option is D
Solution and Explanation
Concept:
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is
\[
K_{\max}=h\nu-\phi
\]
Since the stopping potential $V_s$ is related to the maximum kinetic energy by
\[
eV_s=K_{\max},
\]
we can write
\[
eV_s=h\nu-\phi.
\]
The work function $\phi$ of a photosensitive material is related to its cut-off (threshold) wavelength $\lambda_0$ by
\[
\phi=\frac{hc}{\lambda_0}.
\]
Therefore,
\[
eV_s=\frac{hc}{\lambda}-\frac{hc}{\lambda_0}
\]
or
\[
eV_s=hc\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right).
\]
This relation directly connects the stopping potential with the incident wavelength and the threshold wavelength.
Step 1: Write the equation for the first photosensitive material For the first material, \[ \lambda = 4000\ \text{\AA} \] and \[ \lambda_{01}=4800\ \text{\AA}. \] The stopping potential is given as $V$. Hence, \[ eV = hc\left( \frac{1}{4000} - \frac{1}{4800} \right). \] Taking LCM, \[ eV = hc \left( \frac{6-5}{24000} \right) = \frac{hc}{24000}. \] Thus, \[ eV=\frac{hc}{24000}. \]
Step 2: Write the equation for the second photosensitive material For the second material, \[ \lambda = 4000\ \text{\AA} \] and \[ \lambda_{02}=6000\ \text{\AA}. \] Let the corresponding stopping potential be $V'$. Then, \[ eV' = hc \left( \frac{1}{4000} - \frac{1}{6000} \right). \] Again taking LCM, \[ eV' = hc \left( \frac{3-2}{12000} \right) = \frac{hc}{12000}. \] Therefore, \[ eV'=\frac{hc}{12000}. \]
Step 3: Compare the two stopping potentials Dividing the second equation by the first equation, \[ \frac{eV'}{eV} = \frac{\dfrac{hc}{12000}} {\dfrac{hc}{24000}}. \] The constants $e$, $h$, and $c$ cancel out: \[ \frac{V'}{V} = \frac{24000}{12000} = 2. \] Hence, \[ V' = 2V. \] Therefore, the stopping potential for the second photosensitive material is \[ \boxed{2V}. \]
Step 1: Write the equation for the first photosensitive material For the first material, \[ \lambda = 4000\ \text{\AA} \] and \[ \lambda_{01}=4800\ \text{\AA}. \] The stopping potential is given as $V$. Hence, \[ eV = hc\left( \frac{1}{4000} - \frac{1}{4800} \right). \] Taking LCM, \[ eV = hc \left( \frac{6-5}{24000} \right) = \frac{hc}{24000}. \] Thus, \[ eV=\frac{hc}{24000}. \]
Step 2: Write the equation for the second photosensitive material For the second material, \[ \lambda = 4000\ \text{\AA} \] and \[ \lambda_{02}=6000\ \text{\AA}. \] Let the corresponding stopping potential be $V'$. Then, \[ eV' = hc \left( \frac{1}{4000} - \frac{1}{6000} \right). \] Again taking LCM, \[ eV' = hc \left( \frac{3-2}{12000} \right) = \frac{hc}{12000}. \] Therefore, \[ eV'=\frac{hc}{12000}. \]
Step 3: Compare the two stopping potentials Dividing the second equation by the first equation, \[ \frac{eV'}{eV} = \frac{\dfrac{hc}{12000}} {\dfrac{hc}{24000}}. \] The constants $e$, $h$, and $c$ cancel out: \[ \frac{V'}{V} = \frac{24000}{12000} = 2. \] Hence, \[ V' = 2V. \] Therefore, the stopping potential for the second photosensitive material is \[ \boxed{2V}. \]
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