In a P-type semiconductor, the acceptor energ | 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 acceptor energy level $E_A = 57 \, meV = 57 	imes 10^{-3} \, eV$. Converting this energy int

Vedclass · Accepted Answer

$217100$

Vedclass · Answer

(C) The energy required to create a hole is equal to the acceptor energy level $E_A = 57 \, meV = 57 	imes 10^{-3} \, eV$. Converting this energy into Joules: $E_A = 57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} \, J = 91.2 	imes 10^{-22} \, J$. The relation between energy and wavelength is $E = \frac{hc}{\lambda }$. Therefore, $\lambda = \frac{hc}{E_A} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19}} \, m$. $\lambda = \frac{19.8 	imes 10^{-26}}{91.2 	imes 10^{-22}} \, m = 0.2171 	imes 10^{-4} \, m$. Converting to $\mathring{A}$s: $\lambda = 0.2171 	imes 10^{-4} 	imes 10^{10} \, \mathring{A} = 217100 \, \mathring{A}$.

In a $P$-type semiconductor, the acceptor energy level $E_A$ is located $57 \, meV$ above the valence band. The maximum wavelength required to create a hole is ....... $\mathring{A}$. (Given: $h = 6.6 \times 10^{-34} \, J \, s$, $c = 3 \times 10^8 \, m/s$, $1 \, eV = 1.6 \times 10^{-19} \, J$)

Similar Questions

The electron mobility in $N$-type germanium is $3900 \ cm^2/V \cdot s$ and its conductivity is $6.24 \ mho/cm$. If the effect of holes is negligible,what is the impurity concentration?

Why is it required to add impurity to a pure semiconductor? Mention its condition and explain what impure semiconductors are.

In which of the following statements is the obtained impure semiconductor of $p$-type?

To obtain $P$-type $Si$ semiconductor,we need to dope pure $Si$ with

Doping of a semiconductor (with small impurity atoms) generally changes the resistivity as follows.

Vedclass Test Series

Exam Paper Generator

Online Exam Module