Question

A p-n junction becomes active as photons of wavelength λ=400nm falls on it. Find the energy band gap? 3.09eV 4.51eV 2.45eV

Answer 1

Answer :

Threshold wavelength = 400nm

We have to find the energy band gap of p-n junction$.$

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◈ Energy of photon is given by

• E = hc/λ

h denotes plank constant

c denotes speed of light

λ denotes wavelength

$\leadsto\sf\:E=\dfrac{(6.626\times 10^{-34})(3\times 10^8)}{400\times 10^{-9}}$

$\leadsto\sf\:E=\dfrac{19.878\times 10^{-26}}{400\times 10^{-9}}$

$\leadsto\sf\:E={4.96\times 10^{-19}\:J}$

$\leadsto\sf\:E=\dfrac{4.96\times 10^{-19}}{1.6\times 10^{-19}}$

$\leadsto\boxed{\bf{\red{E=3.09\:eV}}}$

Answer 2

GiveN :

• Wavelength of photons $\sf{\lambda\ =\ 4 \times 10^{-7} m}$

To FinD :

• Energy in band gap.

SolutioN :

We know that,

$\implies \boxed{\boxed{\sf{E\ =\ h \nu}}} \\ \\ \\ \\ \implies \sf{E\ =\ \dfrac{hc}{\lambda}} \\ \\ \\ \\ \implies \sf{E\ =\ \dfrac{(6.626\ \times\ 10^{-34}) \times \: ( 3\ \times\ 10^8)}{4\ \times\ 10^{-7}}} \\ \\ \\ \\ \implies \sf{E\ =\ \dfrac{19.878\ \times\ 10^{-34\ +\ 8}}{4\ \times\ 10^{-7}}} \\ \\ \\ \\ \implies \sf{E\ =\ \dfrac{19.878\ \times\ 10^{-26}}{4\ \times\ 10^{-7}}} \\ \\ \\ \\ \implies \sf{E\ =\ 4.96\ \times\ 10^{-26\ +\ 7}} \\ \\ \\ \\ \implies \sf{E\ =\ 4.96\ \times\ 10^{-19}} \\ \\ \\ \\ \implies \sf{E\ =\ 3.09\ eV}$