Problem — Infinite potential well

Problem

An electron is confined in an infinite potential well of width $L = 0{,}2$ nm. Using $E_n = n^2 h^2/(8 m L^2)$, calculate the energy of the first level ($n = 1$) in eV and comment on the connection with typical atomic energies.

Solution

Step 1 — Formula for the levels. For the infinite well:

$E_n = \frac{n^2 h^2}{8 m L^2}$

with $n = 1$, $L = 0{,}2\;\text{nm} = 2\cdot 10^{-10}$ m, $m = m_e = 9{,}11\cdot 10^{-31}$ kg.

Step 2 — Numerator and denominator.

$h^2 = (6{,}63\cdot 10^{-34})^2 \approx 4{,}40\cdot 10^{-67}\;\text{J}^2\text{s}^2$

$8 m L^2 = 8\cdot 9{,}11\cdot 10^{-31}\cdot (2\cdot 10^{-10})^2 = 8\cdot 9{,}11\cdot 10^{-31}\cdot 4\cdot 10^{-20} \approx 2{,}92\cdot 10^{-49}$

Step 3 — Energy in joules and eV.

$E_1 = \frac{4{,}40\cdot 10^{-67}}{2{,}92\cdot 10^{-49}} \approx 1{,}51\cdot 10^{-18}\;\text{J} = \frac{1{,}51\cdot 10^{-18}}{1{,}6\cdot 10^{-19}} \approx 9{,}4\;\text{eV}$

$\ev{E_1 \approx 9{,}4\;\text{eV}}$

Comment. An electron confined in a region of atomic size ($\sim 0{,}2$ nm) has a ground-level energy of a few eV, the same order as atomic binding energies (the ground state of hydrogen is $13{,}6$ eV). This is not a coincidence: confining the electron imposes, via the uncertainty principle, a minimum momentum and hence a kinetic energy on the electronvolt scale. The infinite well is the simplest model capturing quantisation: only standing waves with an integer number of half-wavelengths across the width $L$ are allowed.

Links

Topics: Fisica quantistica Concepts: Onde stazionarie · Principio di indeterminazione Skills: Impostazione simbolica · Conversione di unità