If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then its band gap energy is :
Planck’s constant h = 6.63 × 10 –34 J.s. Speed of light c = 3 × 10⁸ m/s

Solution

Physics sir · 4 steps

The longest wavelength detected is the photon that just bridges the gap, so set E_g = hc/λₘₐₓ.

1. Why maximum wavelength. A photon is detected only if it can lift an electron across the band gap, so its energy must be at least E_g. Longer wavelength means less energy, and the longest detectable wavelength carries exactly E_g.

   E_g = hc/λₘₐₓ

2. Put in numbers. E_g = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹) = (1.989 × 10⁻²⁵)/(4 × 10⁻⁷) = 4.97 × 10⁻¹⁹ J.
3. Convert to eV. Divide by 1.6 × 10⁻¹⁹ J eV⁻¹: E_g = 3.11 eV. The shortcut E (eV) ≈ 1240/(λ (nm)) = 1240/400 = 3.1 gives the same result.
4. So. The band gap is about 3.1 eV, option (c).

The idea to keep. The cut-off wavelength of a photodiode matches the band gap: E_g = hc/λₘₐₓ. Remember E (eV) ≈ 1240/λ (nm) and these questions take ten seconds.

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