Question

A particle of unknown mass is in a 1-dimensional box of width L = 3.0 x 10-10 m with infinitely high potential walls at x = 0 and at x = L, and zero potential for 0 < x < L. The particle is in the second excited state of the box. What is the de Broglie wavelength, ?, of the particle?

Answer 1

Answer:

0.2 nm

Explanation:

As we know that the energy of the nth level of one dimensional potential box is,

$E_{n}=\frac{h^{2}n^{2}}{8mL^{2} }$

And also it is known that,

$E=\frac{p^{2} }{2m}$

Put this in the above equation.

$\frac{p^{2} }{2m}=\frac{h^{2}n^{2}}{8mL^{2} }\\p=\frac{hn }{2L}$

Now, de Broglie wavelength is,

$\lambda=\frac{h}{p}$

Therefore,

$\lambda=\frac{h2L}{n h } \\\lambda=\frac{2L}{n }$

Given that, the width of 1 dimension box is,

$L=3\times 10^{-10}m$

For the second excited state n=3,

$\lambda=\frac{2\times 3\times 10^{-10}m}{3}\\\lambda=2\times 10^{-10}m\\\lambda=0.2nm$

De Broglie wavelength of the particle is 0.2 nm.