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University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 6: Q40E (page 1355)

For the ground level harmonic oscillator wave function $蠄 (x)$ given in Eq. (40.47), $| 蠄 |^{2}$ has a maximum at x = 0. (a) Compute the ratio of $| 蠄 |^{2}$ at x = +A to $| 蠄 |^{2}$ at x = 0, where A is given by Eq. (40.48) with n = 0 for the ground level. (b) Compute the ratio of $| 蠄 |^{2}$ at x = +2A to $| 蠄 |^{2}$ at x = 0. In each case is your result consistent with what is shown in Fig. 40.27?

Short Answer

Expert verified

(a) The ratio of $| 蠄 |^{2}$ at x = +A to $| 蠄 |^{2}$ at x = 0 is

(b) The ratio of $| 蠄 |^{2}$ at x = +2A to $| 蠄 |^{2}$ at x = 0 is

Step by step solution

01

(a) Determination of the ratio of |ψ|2 at x = +A to |ψ|2 at x = 0.

The wave function in case of a harmonic oscillator is given as,

$蠄 (x) = Cexp (\frac{- mk'}{2 鈩�} x^{2})$

Here C is constant and other symbols have usual meanings.

Now, localid="1664002387765" $蠄 (0) = C Q \exp (0) = 1$

$蠄 (A) = Cexp (\frac{- mk'}{2 鈩�} A^{2})$

So, the ratio of probabilistic ratio is,

$\frac{{|蠄 (A)|}^{2}}{{|蠄 (0)|}^{2}} = \frac{{|Cexp (- \frac{\sqrt{mk'}}{2 鈩�} A^{2})|}^{2}}{{|C|}^{2}}$

localid="1664003260842" $= e x p (- \frac{\sqrt{mk'}}{鈩�} A^{2}) = e x p (- \frac{\sqrt{mk'}}{k'} w^{'})$

$= e x p (- 1) = 0.368$

Thus, the ratio of heat $| 蠄 |^{2}$ at x = +A to $| 蠄 |^{2}$ at x = 0 is 0.368

02

(b) Determination of the ratio of |ψ|2 at x = +2A to |ψ|2 at x = 0

Repeat the above process while replacing A by 2A,

$\frac{{|蠄 (2 A)|}^{2}}{{|蠄 (0)|}^{2}} = \frac{{|Cexp (- \frac{\sqrt{mk'}}{2 鈩�} {(2 A)}^{2})|}^{2}}{{|C|}^{2}}$

$= \exp (- \frac{\sqrt{mk'}}{鈩�} {(4 A)}^{2})$

localid="1664003793111" $= \exp (- \frac{\sqrt{mk'}}{k'} 4 w')$

$= e x p (- 4) = 1.83 脳 10^{- 2}$

Thus, the ratio of $| 蠄 |^{2}$ at x = +2A to $| 蠄 |^{2}$ atx= 0 is $1.83 脳 10^{- 2}$

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