Paragraph: When a particle is restricted to move along $x$-axis between $x=0$ and $x=a$, where $a$ is of…
Paragraph:
When a particle is restricted to move along $x$-axis between $x=0$ and $x=a$, where $a$ is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends $x=0$ and $x=a$. The wavelength of this standing wave is related to the liner momentum $p$ of the particle according to the de Broglie relation. The energy of the particle of mass $m$ is related to its linear momentum as $E=\frac{p^2}{2 m}$. Thus, the energy of the particle can be denoted by a quantum number $n$ taking values $1,2,3, \ldots(n=1$, called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line $x=0$ to $x=a$ [Take $h=6.6 \times 10^{-34} \mathrm{Js}$ and $e=1.6 \times 10^{-19} \mathrm{C}$ ]Question:
If the mass of the particle is $m=1.0 \times 10^{-30} \mathrm{~kg}$ and $a=6.6 \mathrm{~nm}$, the energy of the particle in its ground state is closest to
$0.8 \mathrm{meV}$
$8 \mathrm{meV}$
$80 \mathrm{meV}$
$800 \mathrm{meV}$
Solution
From Eq. (i)
$
E=\frac{n^2 h^2}{8 a^2 m}
$
In ground state $n=1$
$
\therefore \quad E_1=\frac{h^2}{8 m a^2}
$
Substituting the values, we get
$
E_1=8 \mathrm{meV}
$
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