## Related questions with answers

The ground-state energy of an electron trapped in a one-dimensional infinite potential well is 2.6 eV. What will this quantity be if the width of the potential well is doubled?

Solution

VerifiedAn electron confined to an infinite potential well can exist in only certain discrete states. If the well is one-dimensional with length $L$, the energies associated with these quantum states are:

$\begin{align}E_n=\left(\dfrac{h^2}{8mL^2} \right)n^2 \end{align}$

where $m$ is the electron mass and $n$ is a quantum number. Consider an electron in an infinite one-dimensional infinite potential well of length $L$, with a ground state energy of $E_1$, now if the length changed to $L^\prime$, then the new ground state energy is $E_1^\prime$, we need to find the ratio of the $L^\prime$ to $L$, from the first equation we can see for the two ground states everything is constant except the length, so we can write,

$E_1\propto \dfrac{1}{L^2} \qquad E_1^\prime\propto \dfrac{1}{{L^\prime}^2}$

divide these two expression by each others to get,

$\dfrac{{L^\prime}^2}{L^2}=\dfrac{E_1}{E_1^\prime}$

solve for $E^\prime$,to get:

$E_1^\prime=\left( \dfrac{L^2}{{L^\prime}^2}\right) E_1$

if the width of the potential well is doubled then,

$\dfrac{L^\prime}{L}=2 \qquad \rightarrow \qquad \dfrac{L^2}{{L^\prime}^2}=\dfrac{1}{4}$

thus,

$E_1^\prime=\dfrac{E_1}{4}$

The ground-state energy is 2.6 eV, then:

$E_1^\prime=\dfrac{2.6 \mathrm{~eV}}{4}=0.65 \mathrm{~eV}$

$\boxed{E_1^\prime=0.65 \mathrm{~eV}}$

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