Q12P Question: Use the virial theorem... [FREE SOLUTION]

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Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q12P (page 270)

Question: Use the virial theorem (Problem 4.40) to prove Equation 6.55.

Short Answer

Expert verified

It is proved that $<\frac{1}{r}> = \frac{1}{a n^{2}}$ .

Step by step solution

Expression of thepotential energy and Bohr energy

The expression for potential energy is as follows,

$V = \frac{- e^{2}}{4 蟺 {\overset{藱}{o}}_{0}} \frac{1}{r}$

The expression for Bohrenergy is as follows,

localid="1658214362022" $E_{n} = - [\frac{m}{2 h^{2}} {(\frac{- e^{2}}{4 蟺 {\overset{藱}{o}}_{0}})}^{2}] \frac{1}{n^{2}}$

Verification of <1r>=1an2

Write the expression for the relation between potential energy and Bohr energy.

$<V> = 2 E_{n}$

Substitute all the values in the above expression.

localid="1658214341492" $<\frac{- e^{2}}{4 蟺 {\overset{藱}{o}}_{0}} \frac{1}{r}> = - 2 [\frac{m}{2 h^{2}} {(\frac{- e^{2}}{4 蟺 {\overset{藱}{o}}_{0}})}^{2}] \frac{1}{n^{2}} \frac{- e^{2}}{4 蟺 {\overset{藱}{o}}_{0}} <\frac{1}{r}> = - 2 [\frac{m}{2 h^{2}} {(\frac{e^{2}}{4 蟺 {\overset{藱}{o}}_{0}})}^{2}] \frac{1}{n^{2}} <\frac{1}{r}> = \frac{m}{2 h^{2}} {(\frac{e^{2}}{4 蟺 {\overset{藱}{o}}_{0}})}^{2} (\frac{4 蟺 {\overset{藱}{o}}_{0}}{e^{2}}) \frac{1}{n^{2}} = \frac{m}{2 h^{2}} \frac{e^{2}}{4 蟺 {\overset{藱}{o}}_{0}} \frac{1}{n^{2}}$

Consider $a = \frac{4 蟺 {\overset{藱}{o}}_{0} h^{2}}{{me}^{2}}$ , where, a is the Bohr radius.

Substitute $\frac{4 蟺 {\overset{藱}{o}}_{0} h^{2}}{{me}^{2}}$ for a in the above expression.

$<\frac{1}{r}> = \frac{1}{a n^{2}}$

Thus, the equation 6.55 is verified.

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91影视

Question: Use the virial theorem (Problem 4.40) to prove Equation 6.55.

Short Answer

Step by step solution

Expression of thepotential energy and Bohr energy

Verification of <1r>=1an2

One App. One Place for Learning.

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