Question

Electrical conductivity of a semiconductor increases when em radiation of $\lambda <2480nm$ is incident on it. The band gap in (ev) for the semiconductor is

• A. $1.1eV$
• B. $2.5eV$
• C. $0.7eV$
• D. $0.5eV$

Answer 1

Answer: (D)

$0.5eV$

Maximum wavelength required for the electrical conductivity of the semiconductor: ${\lambda }_{max}=2480nm=2480\times {10}^{-9}m$

Thus threshold frequency of the radiation: $\nu =\frac{c}{{\lambda }_{max}}$

Energy band gap for the semiconductor: $E_{gap}=h\nu$

$E_{gap}=6.63\times {10}^{-34}\times \frac{3\times {10}^8}{2480\times {10}^{-9}}J$

$E_{gap}=8.02\times {10}^{-20}J$

$E_{gap}=\frac{8.02\times {10}^{-20}}{1.6\times {10}^{-19}}eV$

$⟹E_{gap}=0.5eV$