Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)

The work function (\(\phi\)) is given in electron volts (eV). To convert it to Joules (J), use the conversion factor 1 eV = \(1.6 \times 10^{-19}\mathrm{J}\). \[\phi = 2.5\,\mathrm{eV} \times \frac{1.6 \times 10^{-19}\mathrm{~J}}{1\mathrm{eV}} = 4.0 \times 10^{-19}\mathrm{J}\]

The energy of the incident light can be calculated using Planck's equation, \(E = h\nu\), where \(E\) is the energy, \(h = 6.62 \times 10^{-34}\mathrm{J.s}\) is Planck's constant, and \(\nu\) is the frequency of the light. First, we need to find the frequency using the speed of light equation, \(c = \lambda\nu\). We are given \(c = 3 \times 10^{8}\,\mathrm{m/s}\) and the wavelength \(\lambda = 3000 \AA = 3000 \times 10^{-10}\,\mathrm{m}\). From the speed of light equation, \[\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8}\,\mathrm{m/s}}{3000 \times 10^{-10}\,\mathrm{m}} = 1.0 \times 10^{15}\,\mathrm{Hz}\] Now we can calculate the energy of the incident light using Planck's equation, \[E = h\nu = (6.62 \times 10^{-34}\mathrm{J.s}) (1.0 \times 10^{15}\,\mathrm{Hz}) = 6.62 \times 10^{-19}\,\mathrm{J}\]

To find the stopping potential (\(V_\text{stop}\)), we will use the photoelectric effect equation, \[V_\text{stop} = \frac{E - \phi}{e}\] where \(E\) is the energy of the incident light calculated in Step 2, \(\phi\) is the work function calculated in Step 1, and \(e = 1.6 \times 10^{-19}\mathrm{C}\) is the elementary charge. Using the values obtained in Step 1 and Step 2, \[V_\text{stop} = \frac{(6.62 \times 10^{-19}\,\mathrm{J}) - (4.0 \times 10^{-19}\,\mathrm{J})}{1.6 \times 10^{-19}\mathrm{C}} = \frac{2.62 \times 10^{-19}\,\mathrm{J}}{1.6 \times 10^{-19}\mathrm{C}}\]

Finally, let's calculate the stopping potential: \[V_\text{stop} = \frac{2.62 \times 10^{-19}\mathrm{J}}{1.6 \times 10^{-19}\mathrm{C}} = 1.64\,\mathrm{V}\] The stopping potential of the emitted electron is \(1.64\,\mathrm{V}\). Thus, the correct answer is (C) \(1.64\,\mathrm{V}\).