Manipal MET2016PhysicsDual Nature of Matter
The wavelength of incident light falling on a photosensitive surface is changed from 2000 Å to 2100 Å . The corresponding change in stopping potential is
Options
- A0.03 V
- B0.3 V
- C3 V
- D3.3 V
Correct answer
B. 0.3 V
Step-by-step solution
Given, aligned ₁=2000 Å & =2000 10⁻¹⁰ ~m & =2 10⁻⁷ ~m aligned ₂=2100 Å=2.1 10⁻⁷ ~m h c ₁ =W+e V₀ (i) and h c ₂ =W+e V₀^ (ii) Subtracting Eq. (ii) from Eq. (i), we get h c ( 1 ₁ - 1 ₂ )=e (V₀-V₀^ ) Change in stopping potential V=V₀-V₀^ aligned & = h c e ( 1 ₁ - 1 ₂ ) & = 6.6 10⁻³⁴ 3 10^8 16 10⁻¹⁹ ( 1 2 10⁻⁷ - 1 2.1 10⁻⁷ ) aligned aligned & = 6.6 3 16 ( 1 2 - 1 2.1 ) & = 6.6 3 0.1 16 2 2.1 =0.3 ~V aligned