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:

• A. $1.5V$
• B. $0.5V$
• C. $4V$
• D. $2V$

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Write the photoelectric relation in wavelengths.
Einstein's equation $eV_s = h\nu - \phi$ with $\phi = \dfrac{hc}{\lambda_0}$ becomes $eV_s = hc\left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right)$.
Step 2: Apply it to the first material.
With $\lambda = 4000\,\text{\AA}$ and $\lambda_{01} = 4800\,\text{\AA}$, $eV = hc\left(\dfrac{1}{4000} - \dfrac{1}{4800}\right) = hc\,\dfrac{6 - 5}{24000} = \dfrac{hc}{24000}$.
Step 3: Apply it to the second material.
With $\lambda = 4000\,\text{\AA}$ and $\lambda_{02} = 6000\,\text{\AA}$, $eV' = hc\left(\dfrac{1}{4000} - \dfrac{1}{6000}\right) = hc\,\dfrac{3 - 2}{12000} = \dfrac{hc}{12000}$.
Step 4: Form the ratio.
$\dfrac{V'}{V} = \dfrac{hc/12000}{hc/24000} = \dfrac{24000}{12000} = 2$.
Step 5: Solve for the new stopping potential.
$V' = 2V$.
Step 6: Conclude.
A material with a longer cut-off wavelength has a smaller work function, so the same photons eject electrons with more energy and a higher stopping potential, $2V$, which is option (4).
\[ \boxed{V' = 2V} \]