Question

Semiconductor Ge has forbidden gap of 1.43 eV. Calculate maximum wavelength which result from electron hole combination.

Answer 1

Correct Answer - 8654 Å
Energy
`E_(min)=(hc)/(lamda_(max))`
`becauselamda_(max)=(hc)/(E_(min))`
`=(6.6xx10^(-34)xx3xx10^(8))/(1.43xx1.6xx10^(-19))m`
`=8.654xx10^(-7)m`
`=8654xx10^(-10)m`
`=8654Å`