Question

The potential for a particle in a one-dimensional box is given as:
$V(x)=0$ for $0≤x≤L,$ and $V(x) = ∞$ elsewhere.
The locations of the internal nodes of the eigenfunctions $\Psi _n(x), n≥2$, are
[Given: m is an integer such that 0<m<n]

• A. $x=\frac{m+\frac1{2}}{n}L$
• B. $x=\frac{m}{n} L$
• C. $x=\frac{m}{n+1} L$
• D. $x=\frac{m+1}{n+1} L$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the locations of the internal nodes of the eigenfunctions \(\Psi_n(x)\) for a particle in a one-dimensional box. The potential for a particle confined in a one-dimensional box is given by:

\(V(x) = 0\) for \(0 \leq x \leq L\), and \(V(x) = \infty\) elsewhere.

The wave functions \(\Psi_n(x)\) of a particle in such a box are given by:

\(\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\) for \(n = 1, 2, 3, \ldots\)

Nodes are the points where the wave function is zero, other than the boundary points. For \(n \geq 2\), the wave function has \((n - 1)\) nodes between \(0\) and \(L\). These nodes occur at points where:

\(\sin\left(\frac{n\pi x}{L}\right) = 0\)

The sine function is zero at integer multiples of \(\pi\), so:

\(\frac{n\pi x}{L} = m\pi\) where \(m\) is an integer such that \(0 < m < n\).

Solving for \(x\), we get:

\(x = \frac{mL}{n}\)

This confirms that the locations of the internal nodes are given by \(x = \frac{m}{n} L\).

Now, let's examine the given options:

• \(x = \frac{m+\frac{1}{2}}{n} L\) - This option is not correct as it does not fit the derived expression.
• \(x = \frac{m}{n} L\) - This is the correct option as it matches with our calculation.
• \(x = \frac{m}{n+1} L\) - This option is incorrect as it does not match the derived expression.
• \(x = \frac{m+1}{n+1} L\) - This option is incorrect as it also does not match the derived expression.

Therefore, the correct answer is \(x = \frac{m}{n} L\).