The photoelectric effect is the phenomenon in which electrons are ejected from the surface of a material when exposed to light of a certain frequency or wavelength. This effect can be explained using the concept of energy quantization. According to Einstein's theory, light can be thought of as particles called photons, and each photon carries a certain amount of energy.

The energy of a photon is given by the equation E = hf, where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), and f is the frequency of the light. The energy required to remove an electron from the surface of a material is known as the work function (Φ) of the material. If the energy of a photon is greater than the work function, the electron will be ejected from the material.

In this question, the radiation used has a wavelength of 250 nm. To find the energy of the photons, we can use the equation c = fλ, where c is the speed of light (3 x 10^8 m/s), f is the frequency, and λ is the wavelength. Rearranging the equation, we get f = c/λ. Substituting the values, we find that the frequency of the radiation is approximately 1.2 x 10^15 Hz.

Now, let's consider the stopping potential applied to stop the ejection of the photoelectron. The stopping potential is the minimum potential difference required to stop the ejected electrons. The stopping potential is related to the maximum kinetic energy (Kmax) of the ejected electrons by the equation Kmax = eV, where e is the charge of an electron (1.6 x 10^-19 C) and V is the stopping potential.

Since the electrons are ejected from the metal, the energy of the photons must be greater than the work function of the metal. Therefore, we can write the equation E = Φ + Kmax, where E is the energy of the photon. Rearranging the equation, we get Kmax = E - Φ.

Since the stopping potential is applied to stop the ejection of the photoelectron, the maximum kinetic energy of the ejected electron will be equal to the energy of the photon (E). Therefore, we can write the equation E = eV.

Now, let's substitute the values into the equations. The energy of the photon is given by E = hf, where h is Planck's constant and f is the frequency. The work function of the metal is given by Φ = hf0, where f0 is the threshold frequency (the minimum frequency required to eject an electron from the metal). We can rearrange this equation to find Φ = (h/f0).

Substituting the values, we get E = hf = (6.626 x 10^-34 J s) x (1.2 x 10^15 Hz) = 7.951 x 10^-19 J. Similarly, Φ = (6.626 x 10^-34 J s) / (f0).

Now, let's find the stopping potential. We know that E = eV, so V = E / e = (7.951 x 10^-19 J) / (1.6 x 10^-19 C) = 4.969 V.

Therefore, the stopping potential required to stop the ejection of the photoelectron is 4.969 V. Since