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 utilized quantum theory to explain the photoelectric effect. He postulated that light comprises photons, each possessing energy: \[ E = hu = \frac{hc}{\lambda} \] Upon interaction with a metal surface, a photon imparts its energy to an electron. If this energy surpasses the material's work function ($ \phi $), electron emission occurs. Einstein's derived equation is: \[ hu = \phi + K_{\text{max}} \Rightarrow K_{\text{max}} = hu - \phi \] Where: - $ h $ denotes Planck's constant. - $ u $ represents the frequency of the incident light. - $ \phi $ signifies the work function. - $ K_{\text{max}} $ is the maximum kinetic energy of the emitted photoelectrons. (b) Step-by-step Analysis: Provided values: \[ \lambda = 330\,\text{nm} = 330 \times 10^{-9}\,\text{m}, \quad h = 6.626 \times 10^{-34}, \quad c = 3 \times 10^8 \] The energy of a single photon is calculated as: \[ E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}} \approx 6.02 \times 10^{-19}\,\text{J} \] Conversion to electronvolts (eV): \[ E = \frac{6.02 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.76\,\text{eV} \] Comparison with Work Functions:

• K: Work function (2.3 eV)<Photon energy (3.76 eV) → Emission will occur.
• Na: Work function (2.75 eV)<Photon energy (3.76 eV) → Emission will occur.
• Mo: Work function (4.17 eV)>Photon energy (3.76 eV) → No emission.
• Ni: Work function (5.15 eV)>Photon energy (3.76 eV) → No emission.

Conclusion: - Metals Molybdenum (Mo) and Nickel (Ni) will not exhibit photoelectric emission. Impact of Decreasing Distance (from 1 m to 0.5 m): - The intensity of the light increases (as intensity is inversely proportional to the square of the distance, $ \propto \frac{1}{r^2} $). - Consequently, the number of emitted electrons increases for metals that are already emitting. - However, the energy of individual photons remains constant. - Therefore, Mo and Ni will still not emit electrons because their work functions exceed the photon energy ($ E<\phi $).