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the potential for a particle in a one dimensional
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]
IIT JAM CY - 2022
IIT JAM CY
Updated On:
Oct 1, 2024
$x=\frac{m+\frac1{2}}{n}L$
$x=\frac{m}{n} L$
$x=\frac{m}{n+1} L$
$x=\frac{m+1}{n+1} L$
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The Correct Option is
B
Solution and Explanation
The correct option is (B): $x=\frac{m}{n} L$
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