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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:
The allowed energy for the particle for a particular value of $n$ is proportional to
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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:
The allowed energy for the particle for a particular value of $n$ is proportional to
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Verified Answer
The correct answer is:
$a^{-2}$
$a^{-2}$

$$
\begin{array}{rlrl}
& a & =\frac{n \lambda}{2} \\
\therefore & \lambda & =\frac{2 a}{n}=\frac{h}{p}=\frac{h}{\sqrt{2 E m}} \\
\text { or } & \sqrt{E} \propto \frac{1}{a} \Rightarrow E \propto \frac{1}{a^2}
\end{array}
$$
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