1. Use Einstein’s photoelectric equation

For incident light of wavelength $\lambda$, the photon energy is

$E = \frac{hc}{\lambda}$

and the maximum kinetic energy is

$K_{\max} = \frac{hc}{\lambda} - \phi$

where $\phi$ is the work function.

2. Write equations for the two cases

• For wavelength $\lambda$:

$K_1 = \frac{hc}{\lambda} - \phi$

• For wavelength $\frac{\lambda}{2}$, photon energy becomes

$\frac{hc}{\lambda/2} = \frac{2hc}{\lambda}$

so

$K_2 = \frac{2hc}{\lambda} - \phi$

3. Use the given condition

It is given that in the second case, maximum kinetic energy is 3 times that in the first case:

$K_2 = 3K_1$

Substitute:

$\frac{2hc}{\lambda} - \phi = 3\left(\frac{hc}{\lambda} - \phi\right)$

4. Solve for $\phi$

Expand the right-hand side:

$\frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi$

Bring like terms together:

$-\frac{hc}{\lambda} + 2\phi = 0$

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

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

5. Match with the options

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

So the correct option is:

B: $\dfrac{hc}{2\lambda}$