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

$A$ hole in a $P-$type semiconductor is

What are $n$-type and $p$-type semiconductors? Name the majority and minority charge carriers in each.

Explain the recombination coefficient of an intrinsic semiconductor and derive the relation $n_i^2 = n_e n_h$.

Assertion: The number of electrons in a $p-$type silicon semiconductor is less than the number of electrons in a pure silicon semiconductor at room temperature.
Reason: It is due to the law of mass action.

The electron concentration in an $n$-type semiconductor is the same as the hole concentration in a $p$-type semiconductor. An external electric field is applied across each of them. Compare the currents in them.

Vedclass Test Series

Exam Paper Generator

Online Exam Module