Answer to Question #127940 in Quantum Mechanics for christopher seebs

Question

 The hydrogen atom wave functions are written as 𝜓𝑛𝑙𝑚. 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 Ψ(𝑟⃗, 0) = 1 √3 𝜓100 + 𝐴𝜓210 where A is a real positive constant. Find A by normalization and determine the wave function at time 𝑡 > 0. Find the average energy of the electron in eV given that the ground state energy is -13.6 eV.

Answer 1

Given wave function is

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

According to Normalization condition,

$\int |\psi(r,0)|^2 dr=1$

$\int |\frac{1}{\sqrt {3}} \psi _{100}+ A \psi_{210}|^2 dr=1$

$\frac{1}{3}\int |\psi _{100}|^2 dr + A^2 \int |\psi _{210}|^2 dr=1$

$\frac{1}{3}+A^2=1$

Since $\int |\psi _{100}|^2 dr =1 and$ $\int |\psi _{210}|^2 dr=1$

On simplifying above equation, we get

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

The wave function as a function of time is

$\psi(r,t) = \frac{1}{\sqrt {3}} \psi _{100} (r,t)+ \sqrt\frac{2}{3} \psi_{210}(r,t)$

$= \frac{1}{\sqrt {3}} \psi _{100} (r)e^{-\frac{iE_1t}{\hbar}}+ \sqrt\frac{2}{3} \psi_{210}(r)e^{-\frac{iE_2t}{\hbar}}$

The average energy of the electron is

$<E>=\frac{1}{3}<E_1>+\frac{2}{3}<E_2>$

$=\frac{1}{3}(-13.6 eV)+\frac{2}{3}(\frac{-13.6 eV}{2^2})$

$=-4.53 eV - 2.27 eV=-6.80 eV$