Question

(a) Briefly explain Einstein’s photoelectric equation.
(b) Four metals with their work functions are listed below:
K = 2.3 eV, Na = 2.75 eV, Mo = 4.17 eV, Ni = 5.15 eV.
The radiation of wavelength 330 nm from a laser source placed 1 m away, falls on these metals.
Which of these metals will not show photoelectric emission?
What will happen if the laser source is brought closer to a distance of 50 cm?

Hint:

Photon energy depends only on wavelength (\( E = \frac{hc}{\lambda} \)), not on distance or intensity. Only metals with \( E \geq \phi \) will emit electrons.

Answer 1

(a) Einstein’s Photoelectric Equation: Einstein explained the photoelectric effect using quantum theory. He proposed that light consists of photons, each with energy: \[ E = h\nu = \frac{hc}{\lambda} \] When a photon strikes a metal surface, it transfers energy to an electron. If this energy exceeds the work function \( \phi \), the electron is emitted. Einstein's equation: \[ h\nu = \phi + K_{\text{max}} \Rightarrow K_{\text{max}} = h\nu - \phi \] where: - \( h \) = Planck’s constant, - \( \nu \) = frequency of incident light, - \( \phi \) = work function, - \( K_{\text{max}} \) = maximum kinetic energy of photoelectrons. (b) Step-by-step Analysis: Given: \[ \lambda = 330\,\text{nm} = 330 \times 10^{-9}\,\text{m}, \quad h = 6.626 \times 10^{-34}, \quad c = 3 \times 10^8 \] Energy of one photon: \[ E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}} \approx 6.02 \times 10^{-19}\,\text{J} \] Convert to eV: \[ E = \frac{6.02 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.76\,\text{eV} \] Compare with work functions: \begin{itemize} \item K = 2.3 eV less than 3.76 → Emission occurs \item Na = 2.75 eV less than 3.76 → Emission occurs \item Mo = 4.17 eV greater than 3.76 → No emission \item Ni = 5.15 eV greater than 3.76 → No emission \end{itemize} Conclusion: - Metals Mo and Ni will not show photoelectric emission. Effect of decreasing distance (from 1 m to 0.5 m): - Intensity increases (since intensity \( \propto \frac{1}{r^2} \)), - Number of emitted electrons increases (for those metals already emitting), - \textit{But} photon energy remains the same, - So Mo and Ni still will not emit electrons (since energy \(<\phi \)).