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Q14P

Expert-verifiedFound in: Page 1215

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An electron is in a certain energy state in a one-dimensional, infinite potential well from x = 0 to x = L =200PM electron’s probability density is zero at x = 0.300 L , and x = 0.400 L ; it is not zero at intermediate values of x. The electron then jumps to the next lower energy level by emitting light. What is the change in the electron’s energy?**

The change in electrons energy or the energy of the photon is

$2.86\times {10}^{-17}\mathrm{J}$ .

**In quantum mechanics, the particle in a box model (also known as the infinite potential well or the infinite square well) describes a particle free to move in a small space surrounded by impenetrable barriers.**

Use the concept of probability density of a practice in an infinite one dimensional box.

The probability density of a wave function of a practice in an infinite one dimensional box of length is given as follows;

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

** **

Here, *L* is the length of the infinite potential well.

The probability density is zero if the angle of sin is integral multiple of $\mathrm{\pi}$ .

${\mathrm{sin}}^{2}\left(\frac{n\mathrm{\pi x}}{L}\right)=0\phantom{\rule{0ex}{0ex}}\frac{n\mathrm{\pi x}}{L}=m\mathrm{\pi}$

This is possible only if is integer that is *m* = 1,2,3,4...

The probability density becomes is not zero between *x* = 0.300 to *x* = 0.400*L* .

Thus is only possible if the state of the electron is *n* = 10.

The next lower energy level is *n'* = 9 if the electron jumps from the state *n* = 10 to the state *n'* = 9 a photon with energy equal the energy difference between the states is emitted.

**$\u2206E=\frac{{n}^{2}{h}^{2}}{8m{L}^{2}}-\frac{{n}^{\text{'}2}{h}^{2}}{8m{L}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{h}^{2}}{8m{L}^{2}}({n}^{2}-{n}^{\text{'}2})$ **

** **

Here, *K* is the Planck’s constant, *m* is the mass of electron, and is the length of the box.

Substitute $6.63\times {10}^{-34}\mathrm{j}.\mathrm{s}\mathrm{for}h,9.11\times {10}^{31}kg\mathrm{for}m\mathrm{for}L,10\mathrm{for}n,\mathrm{and}\mathrm{for}9\mathrm{for}n\text{'}$ .

$\u2206E=\frac{{(6.63\times {10}^{-34}\mathrm{j}.\mathrm{s})}^{2}}{\left(8\right)(9.11\times {10}^{31}\mathrm{kg}){(200\times {10}^{-12}\mathrm{m})}^{2}}({\left(10\right)}^{2}-{\left(9\right)}^{2})=2.86\times {10}^{-17}\mathrm{J}$

Therefore, the change in electrons energy or the energy of the photon is $2.86\times {10}^{-17}\mathrm{J}$.

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