Answer:
Please read the answer below
Explanation:
The energies in a infinite square-well potential of width a is given by
[tex]E=n^2\frac{\pi^2\hbar^2}{8m_ea}[/tex]
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
[tex]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[/tex]
However, the transitions T4-2, T4-3, T3-1 are also allowed
[tex]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[/tex]
hope this helps!!
