The acceptor level of a p-type semiconductor is $6 \mathrm{eV}$. The maximum wavelength of light which can…

The acceptor level of a p-type semiconductor is $6 \mathrm{eV}$. The maximum wavelength of light which can create a hole would be : Given $\mathrm{hc}=1242 \mathrm{eV} \mathrm{nm}$.
  1. $414 \mathrm{~nm}$
  2. $103.5 \mathrm{~nm}$
  3. $207 \mathrm{~nm}$
  4. $407 \mathrm{~nm}$

Solution

$\begin{aligned} & \text { Energy }=\frac{\mathrm{hc}}{\lambda} \\ & \mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV} \\ & 6=\frac{1240}{\lambda(\mathrm{nm})} \\ & \lambda=\frac{1240}{6}=207 \mathrm{~nm}\end{aligned}$

Asked in: JEE Main 2024 (06 Apr Shift 2)

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