A photoelectric surface is illuminated successively by monochromatic light of wavelengths $$\lambda$$ and $$\frac{\lambda}{2}$$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is:

We begin with the photoelectric equation, which relates the maximum kinetic energy $$K_{\text{max}}$$ of the emitted electrons to the frequency $$f$$ of the incident light and the work function $$\phi$$ of the surface:

$$K_{\text{max}} = h f - \phi,$$

where $$h$$ is Planck’s constant.

Next, we convert the given wavelengths into frequencies using the relation $$f = \dfrac{c}{\lambda},$$ where $$c$$ is the speed of light.

For the first illumination, the wavelength is $$\lambda$$. Hence,

$$f_1 = \dfrac{c}{\lambda}.$$

Substituting this frequency into the photoelectric equation, we obtain the first maximum kinetic energy:

$$K_1 = h f_1 - \phi = h\left(\dfrac{c}{\lambda}\right) - \phi = \dfrac{h c}{\lambda} - \phi.$$

Now the surface is illuminated by light of wavelength $$\dfrac{\lambda}{2}$$. The corresponding frequency is

$$f_2 = \dfrac{c}{\lambda/2} = \dfrac{2c}{\lambda}.$$

Again using the photoelectric formula, the second maximum kinetic energy becomes

$$K_2 = h f_2 - \phi = h\left(\dfrac{2c}{\lambda}\right) - \phi = \dfrac{2 h c}{\lambda} - \phi.$$

The problem states that this second kinetic energy is three times the first:

$$K_2 = 3 K_1.$$

Substituting the explicit expressions for $$K_1$$ and $$K_2$$ gives

$$\dfrac{2 h c}{\lambda} - \phi = 3\left(\dfrac{h c}{\lambda} - \phi\right).$$

We now expand the right-hand side:

$$\dfrac{2 h c}{\lambda} - \phi = \dfrac{3 h c}{\lambda} - 3\phi.$$

To isolate $$\phi,$$ we collect like terms. First, bring all terms to one side:

$$0 = \dfrac{3 h c}{\lambda} - 3\phi - \dfrac{2 h c}{\lambda} + \phi.$$

Combine the terms involving $$h c$$ and the terms involving $$\phi$$ separately:

$$0 = \left(\dfrac{3 h c}{\lambda} - \dfrac{2 h c}{\lambda}\right) + (-3\phi + \phi).$$

$$0 = \dfrac{h c}{\lambda} - 2\phi.$$

Now we solve for the work function $$\phi$$ by moving the term with $$\phi$$ to the other side:

$$2\phi = \dfrac{h c}{\lambda}.$$

Finally, dividing by 2 yields

$$\phi = \dfrac{h c}{2\lambda}.$$

This matches Option A in the list provided.

Hence, the correct answer is Option A.