Energy corresponding to threshed frequency of metal is \(6.2 \mathrm{eV}\). If stopping potential corresponding to radiation incident on surface is $5 \mathrm{~V}\(, incident radiation will be in the \)\ldots \ldots \ldots \ldots \ldots$ region. (A) X-ray (B) Ultraviolet (C) infrared (D) Visible

We will use the equation for the photoelectric effect to relate the stopping potential, the threshold frequency energy, and the energy of the incident radiation: \(E_{incident} = E_{threshold} + eV\), where \(E_{incident}\) = energy of incident radiation, \(E_{threshold}\) = energy corresponding to the threshold frequency of the metal, \(e\) = elementary charge (approximately \(1.6 × 10^{−19} \mathrm{C}\)), and \(V\) = stopping potential.

We are given that: \(E_{threshold} = 6.2 \mathrm{eV}\) and \(V = 5 \mathrm{V}\). Using the equation in step 1, we can solve for the energy of the incident radiation: \(E_{incident} = 6.2 \mathrm{eV} + (1.6 × 10^{−19} \mathrm{C})(5 \mathrm{V})\) \(E_{incident} = 6.2 \mathrm{eV} + 8 × 10^{−19} \mathrm{J}\) To keep the same units, let's convert 6.2 eV to Joules: \(6.2 \mathrm{eV} = 6.2 × 1.6 × 10^{−19} \mathrm{J} = 9.92 × 10^{−19} \mathrm{J}\) Now we can find the energy of the incident radiation: \(E_{incident} = 9.92 × 10^{−19} \mathrm{J} + 8 × 10^{−19} \mathrm{J} = 17.92 × 10^{−19} \mathrm{J}\)

Now that we have the energy of the incident radiation, we can use the following equation to find its frequency: \(E = h\nu\), where \(E\) = energy of the radiation, \(h\) = Planck's constant (approximately \(6.626 × 10^{−34} \mathrm{Js}\)), and \(\nu\) = frequency of the radiation. Solving for the frequency: \(\nu = \frac{E}{h} = \frac{17.92 × 10^{−19} \mathrm{J}}{6.626 × 10^{−34} \mathrm{Js}} \approx 2.7 × 10^{15} \mathrm{Hz}\) Next, we'll use the speed of light equation to find the wavelength of the radiation: \(c = \lambda\nu\), where \(c\) = speed of light (approximately \(3 × 10^8 \mathrm{m/s}\)), \(\lambda\) = wavelength of the radiation, and \(\nu\) = frequency of the radiation. Solving for the wavelength: \(\lambda = \frac{c}{\nu} = \frac{3 × 10^8 \mathrm{m/s}}{2.7 × 10^{15} \mathrm{Hz}} \approx 1.1 × 10^{−7}\mathrm{m}\) This value of the wavelength falls within the ultraviolet region (10 − 400 nm), so the correct answer is: (B) Ultraviolet