Question

Radiation of wavelength 200 nm is incident on a photosensitive surface of work function 4·2 eV. The kinetic energy of fastest photoelectrons emitted from this surface will be close to :

• A. 3·5 eV
• B. 3·0 eV
• C. 2·5 eV
• D. 2·0 eV

Hint:

Memorize the value of \(hc\) as \(1240\text{ eV}\cdot\text{nm}\). This constant allows you to convert any wavelength directly into photon energy via simple division, eliminating the need for slow multi-step scientific notation calculations involving \(6.63 \times 10^{-34}\) and \(3 \times 10^8\).

Answer 1

The Correct Option is D

Concept: According to Einstein's Photoelectric Equation, when a photon of energy \(E\) strikes a clean metal surface, its energy is fully absorbed. Part of this energy is used to overcome the binding forces of the electron within the metal latch (known as the work function, \(\phi_0\)), and the remaining portion is converted into the maximum kinetic energy (\(K_{\max}\)) of the escaping electron: \[ E = \phi_0 + K_{\max} \quad \Rightarrow \quad K_{\max} = E - \phi_0 \] The energy \(E\) of an incident photon can be calculated from its wavelength \(\lambda\) using Planck's relation: \[ E = \frac{hc}{\lambda} \] When dealing with atomic scales, using the approximated value of \(hc \approx 1240\text{ eV}\cdot\text{nm}\) simplifies calculations significantly.

Step 1: Calculating the energy of the incident photon in electron-volts (eV).
The wavelength of the incident radiation is given as: \[ \lambda = 200\text{ nm} \] Using the shortcut relation for photon energy: \[ E = \frac{1240\text{ eV}\cdot\text{nm}}{\lambda \text{ (in nm)}} \] Substituting \(\lambda = 200\text{ nm}\): \[ E = \frac{1240}{200} = \frac{124}{20} = 6.2\text{ eV} \] So, each incident photon brings an energy of \(6.2\text{ eV}\) to the metal interface.

Step 2: Calculating the maximum kinetic energy of the emitted photoelectrons.
The work function of the target photosensitive surface is given as: \[ \phi_0 = 4.2\text{ eV} \] Using Einstein's formula to find the maximum kinetic energy: \[ K_{\max} = E - \phi_0 \] Substitute the calculated photon energy and work function: \[ K_{\max} = 6.2\text{ eV} - 4.2\text{ eV} = 2.0\text{ eV} \] The kinetic energy of the fastest photoelectrons emitted is exactly \(2.0\text{ eV}\), matching Option (D).