A P-type semiconductor has acceptor levels 57 | 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 between the valence band and the acceptor level, which is $E = 57 \, meV = 57 	im

Vedclass · Accepted Answer

$217100 \, 	ext{\AA}$

Vedclass · Answer

(C) The energy required to create a hole is equal to the energy gap between the valence band and the acceptor level, which is $E = 57 \, meV = 57 	imes 10^{-3} \, eV$.Converting this energy into Joules: $E = 57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19} \, J$.The relationship between energy and wavelength is given by $E = \frac{hc}{\lambda}$.Therefore, the maximum wavelength $\lambda$ is $\lambda = \frac{hc}{E}$.Substituting the values: $\lambda = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{57 	imes 10^{-3} 	imes 1.6 	imes 10^{-19}} \, m$.$\lambda = \frac{19.8 	imes 10^{-26}}{91.2 	imes 10^{-22}} \, m \approx 0.2171 	imes 10^{-4} \, m$.Converting to $	ext{\AA}$s: $\lambda = 0.2171 	imes 10^{-4} 	imes 10^{10} \, 	ext{\AA} = 217100 \, 	ext{\AA}$.

$A$ $P$-type semiconductor has acceptor levels $57 \, meV$ above the valence band. The maximum wavelength of light required to create a hole is (Planck's constant $h = 6.6 \times 10^{-34} \, J-s$, speed of light $c = 3 \times 10^8 \, m/s$, charge of electron $e = 1.6 \times 10^{-19} \, C$)

Similar Questions

The valence of an impurity added to germanium crystal in order to convert it into a $P-$type semiconductor is

When a small amount of antimony impurity is added to a germanium crystal,what happens?

In extrinsic $P$ and $N$-type semiconductor materials,the ratio of the impurity atoms to the pure semiconductor atoms is about

In $N$-type semiconductors,the majority charge carriers are

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module