Answer to Question #216139 in Molecular Physics | Thermodynamics for GOPINATH K

Question

Calculate the probability of finding the particle
between x= 0 to x=L/100, in an infinite potential
well of 10 Angstrom width. Given that the particle is in 100 th excited state.

Answer 1

At first, we know that for a particle in a box:

$\Psi_{n}(x)=\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L}), n=1, 2, 3, ...$

For the problem, we have n=101 (because n=2 is the first excited state), and we will calculate the probability with the integral

$P=\int_0^{L/100}[\Psi_{n}(x)]^2dx \\ P=\int_0^{L/100}[\sqrt{\frac{2}{L}}sin(\frac{n\pi x}{L})]^2dx \\ P=\frac{2}{L}\int_0^{L/100}sin^2(\frac{n\pi x}{L})dx$

Then, to solve the integral we use the formula for the squared sine:

$sin^2(\frac{n\pi x}{L})=\frac{1}{2}[1-cos(\frac{2n\pi x}{L})]$

We substitute and find the result of the integral:

$P=\frac{2}{L}\int_0^{L/100}sin^2(\frac{n\pi x}{L})dx \\ P=\frac{1}{L}\int_0^{L/100}[1-cos(\frac{2n\pi x}{L})]dx \\P= \frac{1}{L} [x-\frac{L}{2n\pi}sin(\frac{2n\pi x}{L})]\lvert_0^{L/100}$

Then we substitute the limits to find the probability between x=0 and x=L/100

$P= \frac{1}{L}\{ [\frac{L}{100}-\frac{L}{2(101)\pi}sin(\frac{2\pi(101) \frac{L}{100}}{L})]-[0-\frac{L}{2(101)\pi}sin(0)]\} \\P= \frac{1}{L}\{\frac{L}{100}-\frac{L}{202\pi}sin(2.02\pi)\}$

This last formula after simplification gives $\\ P= \frac{1}{100}-\frac{sin(2.02\pi)}{202\pi}$ , which is equal to

$P= \frac{1}{100}-\frac{sin(2.02\pi)}{202\pi}\approxeq0.01-\frac{0.06279}{202\pi}\approxeq0.01-0.000098945$

$\\ P\approxeq 0.009901; \% P\approxeq 0.9901 \%$

In conclusion, the probability to find the particle between x= 0 to x=L/100, in an infinite potential

well given that the particle is in the 100th excited state is P=0.009901 or 0.9901%.

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.