1. Use Einstein’s photoelectric equation

For incident wavelength $\lambda$, the stopping potential $V_0$ satisfies

$eV_0 = \frac{hc}{\lambda} - \phi$

where $\phi$ is the work function of the metal.

For incident wavelength $2\lambda$, the stopping potential becomes $\dfrac{V_0}{4}$, so

$e\left(\frac{V_0}{4}\right) = \frac{hc}{2\lambda} - \phi$

2. Write both equations clearly

From the first case:

$eV_0 = \frac{hc}{\lambda} - \phi \qquad (1)$

From the second case:

$\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi \qquad (2)$

3. Eliminate $V_0$ to find $\phi$

From (1), divide by $4$:

$\frac{eV_0}{4} = \frac{hc}{4\lambda} - \frac{\phi}{4} \qquad (3)$

Now compare (2) and (3):

$\frac{hc}{4\lambda} - \frac{\phi}{4} = \frac{hc}{2\lambda} - \phi$

Multiply throughout by $4$:

$\frac{hc}{\lambda} - \phi = \frac{2hc}{\lambda} - 4\phi$

Rearrange:

$3\phi = \frac{hc}{\lambda}$

so,

$\phi = \frac{hc}{3\lambda}$

4. Find the threshold wavelength

Threshold wavelength $\lambda_0$ is given by

$\phi = \frac{hc}{\lambda_0}$

Substitute $\phi = \dfrac{hc}{3\lambda}$:

$\frac{hc}{\lambda_0} = \frac{hc}{3\lambda}$

Hence,

$\lambda_0 = 3\lambda$

5. Check options

• A: $3\lambda$ ✅
• B: $4\lambda$
• C: $\dfrac{3}{2}\lambda$
• D: $\dfrac{\lambda}{4}$

So the correct option is A.