Question

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency of the material is
1. 1.6 × 10¹⁵ Hz
2. 2.5 × 10¹⁵ Hz
3. 4 × 10¹⁵ Hz
4. 5 × 10¹⁵ Hz

Answer 1

Correct Answer - Option 1 : 1.6 × 10¹⁵ Hz

CONCEPT : 

• The energy emitted when the atom jump from a higher energy level to a lower energy level is given by

\(⇒ E = 13.6 [ \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]\)

Where n_i = Initial energy level, n_f =Final energy level

• Stopping potential is the minimum negative potential that applied to stop the emission of photoelectrons 
• Stopping potential is  the maximum kinetic energy of an  emitted photoelectron 

CALCULATION :

Given  - Stopping potential = 3.57 V, KEmax = 3.57 eV, n_i = 1 , n_f = 2

• The energy emitted when the atom jump from a higher energy level to a lower energy level is given by

\(⇒ E = 13.6 [ \frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}]\)

Substituting the given values in the above equation it becomes 

\(⇒ E= 13.6 \left[\dfrac{1}{1^2} - \dfrac{1}{2^2}\right]=3.4 × 3 = 10.2 \ eV\)

The incident emrgy can be treated as the kinetic energy , hence KE = 10.2 eV

⇒ W₀ = 10.2 eV - 3.57 eV or W₀ = 6.63 eV

\(\Rightarrow v = \dfrac{6.63 × 1.6 × 10^{-19}}{h}\) or v = 1.6 × 10¹⁵

• Hence, option 1 is the answer