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Q5Q

Expert-verifiedFound in: Page 1214

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A proton and an electron are trapped in identical one-dimensional infinite potential wells; each particle is in its ground state. At the center of the wells, is the probability density for the proton greater than, less than, or equal to that of the electron?**

The probability density for the proton is equal to that of the electron.

The given data can be listed below as,

- The location of each particle is,
*n*=1 (ground state).

**The term ‘probability density’ does have a physical meaning. It represents the probability that the electron will be detected in a specific interval. If the probability density of a charged particle is integrated over an entire axis means the total probability must be equal to 1.**

The expression of the probability density of a charged particle can be expressed as,

** **

${\psi}_{n}^{2}\left(x\right)=\left(\frac{2}{L}\right){\mathrm{sin}}^{2}\left(\frac{n\mathrm{\pi}}{L}x\right)$

** **

Here,${\psi}_{n}^{2}\left(x\right)$is the probability density of a charged particle and L is the length of the well.

Substitute the value in the above expression.

${\psi}_{n}^{2}\left(x\right)=\left(\frac{2}{L}\right){\mathrm{sin}}^{2}\left[\frac{\left(1\right)\mathrm{\pi}}{L}x\right]\phantom{\rule{0ex}{0ex}}=\left(\frac{2}{L}\right){\mathrm{sin}}^{2}\left[\frac{\mathrm{\pi}}{L}x\right]$

From the above expression, one can observe that the expression of the probability density does not have any variable/parameter related to the different types of charged particles. This means the expression of the probability density for the proton and electron would be the same.

** **

Thus, the probability density for the proton is equal to that of the electron.

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