Question

The acceptor level of a p-type semiconductor is 6eV. The maximum wavelength of light which can create a hole would be : Given hc = 1242 eV nm. 

(1) 407 nm 

(2) 414 nm 

(3) 207 nm 

(4) 103.5 nm

Answer 1

Correct option is : (3) 207 nm

Energy \(=\frac{\mathrm{hc}}{\lambda}\);

\(\mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV}\)

\(6=\frac{1240}{\lambda(\mathrm{nm})}\)

\(\lambda=\frac{1240}{6}=207 \mathrm{nm}\)