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Alexeev081 [22]
3 years ago
14

An electron is trapped in an infinite square-well potential of width 0.2 nm. If the electron is initially in the n = 4 state, wh

at are the various photon energies that can be emitted as the electron jumps to the ground state? (List in descending order of energy. Enter 0 in any remaining unused boxes.)
Physics
1 answer:
eimsori [14]3 years ago
6 0

Answer:

Please read the answer below

Explanation:

The energies in a infinite square-well potential of width a is given by

E=n^2\frac{\pi^2\hbar^2}{8m_ea}

where me=9.1*10^{-31}kg, hbar=1.05*10^{-34}Js and a= 0.2*10^{-9}m.

From the state n=4 the electron can pass to state n=3, n=2 and n=1. The different transitions can be

n4->n3=E4-E3

n3->n2=E3-E2

n2->n1=E2-E1

Hence, by replacing we have that the photon energies emitted are given by

T_{4-3}=E_4-E_3=\frac{\pi^2\hbar^2}{8m_ea}(4^2-3^2)=5.26*10^{-28}J\\T_{3-2}=E_3-E_2=\frac{\pi^2\hbar^2}{8m_ea}(3^2-2^2)=3.76*10^{-28}J\\T_{2-1}=E_2-E_1=\frac{\pi^2\hbar^2}{8m_ea}(2^2-1^2)=2.25*10^{-28}J

However, the transitions T4-2, T4-3, T3-1 are also allowed

T_{4-2}=E_4-E_2=\frac{\pi^2\hbar^2}{8m_ea}(4^2-2^2)=9.04*10^{-28}J\\T_{4-1}=E_4-E_1=\frac{\pi^2\hbar^2}{8m_ea}(4^2-1^2)=1.12*10^{-27}J\\T_{3-1}=E_3-E_1=\frac{\pi^2\hbar^2}{8m_ea}(3^2-1^2)=6.01*10^{-28}J

hope this helps!!

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Explanation:

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1 / f = 1 / di + 1 / do

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Then,

1 / f_B = 1 / di_A + 1 / do

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28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

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Distance of object from the lens when Billy uses Annie glass

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