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# 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?

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An 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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