Question #127940

 The hydrogen atom wave functions are written as πœ“π‘›π‘™π‘š. State the values of n, l, and m. State the relation between a physical quantity and each quantum number. At t = 0 the hydrogen atom is in the superposition state Ξ¨(π‘Ÿβƒ—, 0) = 1 √3 πœ“100 + π΄πœ“210 where A is a real positive constant. Find A by normalization and determine the wave function at time 𝑑 > 0. Find the average energy of the electron in eV given that the ground state energy is -13.6 eV.


1
Expert's answer
2020-08-04T16:00:38-0400

Given wave function is

ψ(r,0)=13ψ100+Aψ210\psi(r,0) = \frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}

According to Normalization condition,

∫∣ψ(r,0)∣2dr=1\int |\psi(r,0)|^2 dr=1

∫∣13ψ100+Aψ210∣2dr=1\int |\frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}|^2 dr=1

13∫∣ψ100∣2dr+A2∫∣ψ210∣2dr=1\frac{1}{3}\int |\psi _{100}|^2 dr + A^2 \int |\psi _{210}|^2 dr=1

13+A2=1\frac{1}{3}+A^2=1

Since ∫∣ψ100∣2dr=1and\int |\psi _{100}|^2 dr =1 and ∫∣ψ210∣2dr=1\int |\psi _{210}|^2 dr=1

On simplifying above equation, we get

A=23A=\sqrt\frac{2}{3}

The wave function as a function of time is

ψ(r,t)=13ψ100(r,t)+23ψ210(r,t)\psi(r,t) = \frac{1}{\sqrt {3}} \psi _{100} (r,t)+ \sqrt\frac{2}{3} \psi_{210}(r,t)

=13ψ100(r)eβˆ’iE1tℏ+23ψ210(r)eβˆ’iE2tℏ= \frac{1}{\sqrt {3}} \psi _{100} (r)e^{-\frac{iE_1t}{\hbar}}+ \sqrt\frac{2}{3} \psi_{210}(r)e^{-\frac{iE_2t}{\hbar}}

The average energy of the electron is

<E>=13<E1>+23<E2><E>=\frac{1}{3}<E_1>+\frac{2}{3}<E_2>

=13(βˆ’13.6eV)+23(βˆ’13.6eV22)=\frac{1}{3}(-13.6 eV)+\frac{2}{3}(\frac{-13.6 eV}{2^2})

=βˆ’4.53eVβˆ’2.27eV=βˆ’6.80eV=-4.53 eV - 2.27 eV=-6.80 eV


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