Q.30 State Einstein’s photoelectric equation and mention physical significance of each term involved in it. The wavelength of incident light is 4000Å. Calculate the energy of incident photon.
Best answer
Answer: \(h\nu = \phi_0 + \frac{1}{2}mv_{max}^2\); energy of photon \(\approx 4.97 \times 10^{-19}\) J (about 3.1 eV).
Einstein’s photoelectric equation:
\[h\nu = \phi_0 + \frac{1}{2}mv_{max}^2 \quad\text{or}\quad h\nu = h\nu_0 + K_{max}\]Significance of terms:
The equation explains why there is no emission below \(\nu_0\) and why \(K_{max}\) depends linearly on frequency but not on intensity.
Numerical: \(\lambda = 4000\) Å \(= 4 \times 10^{-7}\) m.
\[E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4 \times 10^{-7}} = 4.97 \times 10^{-19} \text{ J}\]\[E = \frac{4.97 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.1 \text{ eV}\]Log in or create a free account to answer this question.
