Light source having wavelength \(331\) nm is used to generate photo-electrons whose stopping potential is \(0.2\) V. The work function of the used metal in the experiment is \(\alpha \times 10^{-19}\) J. The value of \(\alpha\) is _____.
(\(h = 6.62 \times 10^{-34}\) J s, \(e = 1.6 \times 10^{-19}\) C and \(c = 3 \times 10^8\) m/s)

Using Einstein's photoelectric equation:

\(\dfrac{hc}{\lambda} = \Phi + eV_0\)

The energy of the incident photon is:

\(E = \dfrac{hc}{\lambda} = \dfrac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}\)

\(E = \dfrac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 6 \times 10^{-19}\) J

The maximum kinetic energy of the photoelectrons is:

\(K_{\text{max}} = eV_0 = 1.6 \times 10^{-19} \times 0.2 = 0.32 \times 10^{-19}\) J

The work function \(\Phi\) is:

\(\Phi = E - K_{\text{max}}\)

\(\Phi = 6 \times 10^{-19} - 0.32 \times 10^{-19} = 5.68 \times 10^{-19}\) J

Comparing with \(\alpha \times 10^{-19}\) J, we get \(\alpha = 5.68\).

Answer: \(5.68\)