Question

What is the total probability of finding a particle in a one-dimensional box at level$\mathbf{n}\mathbf{=}\mathbf{3}$between$\mathbf{x}\mathbf{=}\mathbf{0}$and$\mathbf{x}\mathbf{=}\mathbf{L}\mathbf{/}\mathbf{6}$?

Short answer

The total probability to find the particle in a one-dimensional box at level between$\mathrm{x}=0$ and$\mathrm{x}=\mathrm{L}/6$ is 0.167.

Step-by-step solution

Boundary conditions for particles present in a one-dimensional box.

Below mentioned are boundary conditions for particles present in a one-dimensional box.

1. Particle is always bounded to be present inside the box only and it cannot be a present outside box.

2. Total probability to find the particle in the box is always one.

3. Wave function must be continuous.

Find the total probability to find the particle in the 1D-box.

Expression for the wave function of a particle in a one-dimensional box is as follows:

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

Where,

$\mathrm{\psi }\left(\mathrm{x}\right)$- is wave function for a particle in a one-dimensional box.

-L is length of the one-dimensional box.

-n is an integer.

-x is an arbitrary point on box length.

Expression for probability to find the electron in the one-dimensional box is as follows:

Total probability $={\int }_0^{\mathrm{L}}{\mathrm{\psi }}^2\left(\mathrm{x}\right)\mathrm{dx}$

Value of $\mathrm{\psi }\left(\mathrm{x}\right)$is$\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)$

Substitute the values in the above equation.

Total probability$={\int }_0^{\mathrm{L}}\sqrt{\frac{2}{\mathrm{L}}}{\sin }^2\left(\frac{\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)$

The above equation is to be integrated between$\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}/6$as follows:

Total probability$={\int }_0^{\mathrm{L}/6}\sqrt{\frac{2}{\mathrm{L}}}{\sin }^2\left(\frac{\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)$

The above equation is integrated as follows:

Total probability

$$\begin{gathered} =\frac{2}{\mathrm{L}}{\int }_0^{\mathrm{L}}1-\cos 2\left(\frac{\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)\mathrm{dx} \\ =\left(\frac{2}{\mathrm{L}}\right)\left(\frac{1}{2}\right){\left[1\mathrm{dx}-\cos 2\left(\frac{\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)\mathrm{dx}\right]}_0^{\mathrm{L}/6} \\ =\frac{1}{\mathrm{L}}{\left[\mathrm{x}-\left(\frac{1}{2\mathrm{n\pi }}\right)\sin \left(\frac{2\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)\mathrm{dx}\right]}_0^{\mathrm{L}/6} \end{gathered}$$

Value of is 3.

The value of the lower limit is 0.

Value of upper limit is $\mathrm{L}/6$.

Substitute the values in the above equation.

Total probability

$$\begin{gathered} =\frac{1}{\mathrm{L}}{\left[\mathrm{x}-\left(\frac{1}{2\mathrm{n\pi }}\right)\sin \left(\frac{2\mathrm{n\pi }}{\mathrm{L}}\mathrm{x}\right)\mathrm{dx}\right]}_0^{\mathrm{L}/6} \\ =\frac{1}{\mathrm{L}}\left[\left(\frac{\mathrm{L}}{6}-0\right)-\left(\frac{1}{\left(2\right)\left(3\right)\mathrm{\pi }}\right)\sin \left(\frac{\left(2\right)\left(3\right)\mathrm{\pi }}{\mathrm{L}}\left(\frac{\mathrm{L}}{6}-0\right)\right)\right] \\ =\frac{1}{\mathrm{L}}\left[\left(\frac{\mathrm{L}}{6}\right)-\left(\frac{1}{6\mathrm{\pi }}\right)\sin \left(\frac{6\mathrm{\pi }}{\mathrm{L}}\left(\frac{\mathrm{L}}{6}\right)\right)\right] \\ =\frac{1}{6} \end{gathered}$$

Hence, the required probability is$\frac{1}{6}$ or 0.167.

Conclusion

Total probability to find the particle in a one-dimensional box in level$\mathrm{n}=3$ between$\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}/6$is $0.167$.