Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm. The band gap of photodiode is found to be (X/8) eV. The value of X is :
(Given, h=6.6 × 10⁻³⁴ Js, e=1.6 × 10⁻¹⁹ C )

Solution

Physics sir · 4 steps

The cutoff wavelength gives the smallest photon energy that works, and that energy is the band gap.

1. Why the cutoff. Conductivity changes only when a photon lifts an electron from the valence band to the conduction band. So the photon energy must be at least E_g. The longest wavelength that works, 660 nm, carries exactly E_g.

   E_g = hc/λ = (6.6 × 10⁻³⁴ × 3 × 10⁸)/(660 × 10⁻⁹) J

2. Put in numbers. The numerator is 19.8 × 10⁻²⁶ and the denominator is 6.6 × 10⁻⁷, so E_g = 3 × 10⁻¹⁹ J.
3. Convert to eV. Divide by 1.6 × 10⁻¹⁹: E_g = 3/1.6 eV = 1.875 eV. Since 1.875 = 15/8, this matches X/8 with X = 15.
4. So. X = 15, which is option (c).

The idea to keep. For a photodiode, the threshold wavelength sets the band gap: E_g = hc/λₘₐₓ. A quick check is E (eV) ≈ 1240/λ (nm).

Answer from the NTA final answer key. Solution by PrepWiser.