Question

Forbidden energy gap of $\text{Ge is}0.75\text{eV}$, maximum wavelength of incident radiation for producing electron-hole pair in germanium semiconductor is-
[Use $h=6.6\times {10}^{-34}\text{Js}$]

• A. $4200\mathring{A}$
• B. $16500\mathring{A}$
• C. $4700\mathring{A}$
• D. $4000\mathring{A}$

Answer 1

The correct option is B $16500\mathring{A}$

Given, $E_G=0.75\text{eV}$

$E_G=\frac{hc}{\lambda }$

$h=6.6\times {10}^{-34}\text{Js};c=3\times {10}^8{\text{ms}}^{-1}$

${\lambda }_{max}=\frac{hc}{E_G}$

$=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{0.75\times 1.6\times {10}^{-19}}=16.5\times {10}^{-7}\text{m}\text{(or)}$

$=16500\mathring{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.