Consider a situation in which reserve biased current of a particular P-N junction increases when it is exposed to a light of wavelength ≤ 621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly.

Solution

Physics sir · 3 steps

The cut-off wavelength is the photon that just matches the band gap, so turn 621 nm into eV.

1. Why a threshold. A photon makes a hole-electron pair only if its energy is at least the band gap E_g. The reverse current rises for λ ≤ 621 nm, so the longest wavelength that works, 621 nm, carries exactly E_g.

   E_g = hc/λₘₐₓ = (1240 eV nm)/(621 nm)

2. Put in numbers. 1240/621 ≈ 2.0. In SI units you get the same result: (6.63 × 10⁻³⁴ × 3 × 10⁸)/(621 × 10⁻⁹) = 3.2 × 10⁻¹⁹ J, and dividing by 1.6 × 10⁻¹⁹ gives 2.0 eV.
3. So. The band gap is nearly 2 eV, option (d).

The idea to keep. Photo-generation in a diode has a threshold: the largest wavelength that still works gives the band gap through E_g = 1240/λ (with λ in nm and E_g in eV). Remember hc = 1240 eV nm.

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