The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72 eV)

Question

The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72 eV) (A) 1.7 × 10-6m (B) 1 .5 × 10 -5m (C) 1.3 × 10-4 m (D) 1.9 × 10-5 m

Tardigrade Admin · Accepted Answer

Here, Eg = 0.72 eV = 0.72 × 1.6 × 10-19 J If λ is the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium, then Eg = (hc/λ) or λ = (hc/Eg) =(6.62× 10-34× 3× 108/0.72× 1.6×10-19) = 1.7× 10-6m

Q. The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is $0.72 e V$ )

$1.7 \times 1 0^{- 6} m$

$1.5 \times 1 0^{- 5} m$

$1.3 \times 1 0^{- 4} m$

$1.9 \times 1 0^{- 5} m$

Q. The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is 0.72eV)

1.7×10−6m

1.5×10−5m

1.3×10−4m

1.9×10−5m

Here, Eg​=0.72eV =0.72×1.6×10−19J If λ is the maximum wavelength of electromagnetic radiation which can create a hole-electron pair in germanium, then Eg​=λhc​ or λ=Eg​hc​ =0.72×1.6×10−196.62×10−34×3×108​ =1.7×10−6m

Q. The maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. (Given that forbidden energy gap in germanium is $0.72 e V$ )

$1.7 \times 1 0^{- 6} m$

$1.5 \times 1 0^{- 5} m$

$1.3 \times 1 0^{- 4} m$

$1.9 \times 1 0^{- 5} m$