Explanation:
Photoelectric effect is the phenomenon where electrons are emitted from a metal surface when light of certain frequency or wavelength falls on it. In this question, we are given that the stopping potential for a given metal is V0 when radiation of wavelength λ0 is used. We need to find the stopping potential when radiation of wavelength 2 λ0 is used.

The stopping potential for photoelectric effect depends on the frequency or wavelength of the incident radiation and the work function of the metal. The work function is the energy required to remove an electron from the metal surface.

The stopping potential is defined as the potential difference between the metal surface and the collector electrode that just stops the photoelectrons from reaching the collector. When the stopping potential is applied, the photoelectrons are brought to rest and their kinetic energy is converted to electrical potential energy.

We can use the following equation to relate the stopping potential with the frequency or wavelength of the incident radiation and the work function of the metal:

eV0 = hf - Φ

where e is the charge of an electron, V0 is the stopping potential, h is Planck's constant, f is the frequency of the incident radiation, λ is the wavelength of the incident radiation, and Φ is the work function of the metal.

Let's first consider the case when radiation of wavelength λ0 is used. From the given information, we have:

eV0 = hc/λ0 - Φ

where c is the speed of light in vacuum.

Now, let's consider the case when radiation of wavelength 2 λ0 is used. The frequency of the radiation is halved, so we have:

f = c/2 λ0

Using the same equation as above, we can write:

eV = hf - Φ = hc/2 λ0 - Φ

Dividing this equation by the first equation, we get:

V/V0 = (hc/2 λ0 - Φ)/ (hc/ λ0 - Φ)

Simplifying this equation, we get:

V = V0 (1 - λ0/2 λ0)

V = V0 (1/2)

Therefore, the stopping potential when radiation of wavelength 2 λ0 is used is V0/2, which is option (a). However, this is not one of the given options. The closest option is (d), which is V0 - hc/2eλ0. We can write this equation as:

V = V0 - hc/2eλ0

This equation implies that the stopping potential is reduced when the wavelength of the incident radiation is doubled. This is consistent with the fact that the frequency of the radiation is halved, so the stopping potential should decrease. Therefore, option (d) is the correct answer.