The acceptor level of a p-type semiconductor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy required to create a hole is equal to the energy gap of the acceptor level,$E = 6 \ eV$.The relationship between energy and wavelength

Vedclass · Accepted Answer

$207$

Vedclass · Answer

(C) The energy required to create a hole is equal to the energy gap of the acceptor level,$E = 6 \ eV$.The relationship between energy and wavelength is given by $E = \frac{hc}{\lambda}$.Given $hc = 1242 \ eV \ nm$,we can write:$6 \ eV = \frac{1242 \ eV \ nm}{\lambda}$Solving for $\lambda$:$\lambda = \frac{1242}{6} \ nm$$\lambda = 207 \ nm$.Therefore,the maximum wavelength of light that can create a hole is $207 \ nm$.

The acceptor level of a $p$-type semiconductor is $6 \ eV$ above the valence band. The maximum wavelength of light which can create a hole would be: (Given $hc = 1242 \ eV \ nm$) (in $nm$)

Similar Questions

If $n_{e}$ and $n_{h}$ are concentrations of electrons and holes in a semiconductor,then the intrinsic carrier concentration $(n_i)$ in thermal equilibrium is

Which of the following energy band diagrams shows the $N$-type semiconductor?

Which one of the following statements is false?

Which of the following statements is true for an $N$-type semiconductor?

In an $n$-type silicon,which of the following statements is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module