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 acceptor level and the valence band,which is $E = 57 \, meV = 57 	ime

Vedclass · Accepted Answer

$217100 \, \mathring{A}$

Vedclass · Answer

(C) The energy required to create a hole is equal to the energy gap between the acceptor level and the valence band,which is $E = 57 \, meV = 57 	imes 10^{-3} \, eV$.Using the relation between energy in $eV$ and wavelength in $\mathring{A}$:$E = \frac{12400}{\lambda (\mathring{A})}$Rearranging for wavelength:$\lambda = \frac{12400}{E(eV)} = \frac{12400}{57 	imes 10^{-3}} \, \mathring{A}$$\lambda \approx 217543 \, \mathring{A}$.Rounding to the nearest provided option,the correct value is $217100 \, \mathring{A}$.

$P$-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 $h = 6.6 \times 10^{-34} \, J-s$)

Similar Questions

$A$ $P-$ type semiconductor can be obtained by adding

In the energy band diagram of a material shown in the figure,the open circles and filled circles denote holes and electrons,respectively. The material is:

Which one of the following statements is false?

Doping of intrinsic semiconductor is done:

An $n$-type and $p$-type semiconductor can be obtained by respectively doping pure silicon with

Vedclass Test Series

Exam Paper Generator

Online Exam Module