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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 LL, the energies associated with these quantum states are:

En=(h28mL2)n2\begin{align}E_n=\left(\dfrac{h^2}{8mL^2} \right)n^2 \end{align}

where mm is the electron mass and nn is a quantum number. Consider an electron in an infinite one-dimensional infinite potential well of length LL, with a ground state energy of E1E_1, now if the length changed to LL^\prime, then the new ground state energy is E1E_1^\prime, we need to find the ratio of the LL^\prime to LL, from the first equation we can see for the two ground states everything is constant except the length, so we can write,

E11L2E11L2E_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,

L2L2=E1E1\dfrac{{L^\prime}^2}{L^2}=\dfrac{E_1}{E_1^\prime}

solve for EE^\prime,to get:

E1=(L2L2)E1E_1^\prime=\left( \dfrac{L^2}{{L^\prime}^2}\right) E_1

if the width of the potential well is doubled then,

LL=2L2L2=14\dfrac{L^\prime}{L}=2 \qquad \rightarrow \qquad \dfrac{L^2}{{L^\prime}^2}=\dfrac{1}{4}

thus,

E1=E14E_1^\prime=\dfrac{E_1}{4}

The ground-state energy is 2.6 eV, then:

E1=2.6 eV4=0.65 eVE_1^\prime=\dfrac{2.6 \mathrm{~eV}}{4}=0.65 \mathrm{~eV}

E1=0.65 eV\boxed{E_1^\prime=0.65 \mathrm{~eV}}

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