Question

When a particle is restricted to move along the 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 linear 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 \[E=\dfrac{{{p}^{2}}}{2m}\]. Thus, the energy of the particle can be denoted by a quantum number ‘\[n\]’ taking values \[1,2,3...\] (\[n=1\], called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following question for a particle moving in the line \[x=0\] to \[x=a\]. Take \[h=6.6\times {{10}^{-34}}Js\] and \[e=1.6\times {{10}^{-19}}c\].
The allowed energy for the particle for a particular value of n is proportional to
A)\[{{a}^{-2}}\]
B) \[{{a}^{-\dfrac{3}{2}}}\]
C) \[{{a}^{-1}}\]
D) \[{{a}^{2}}\]

Answer 1

Hint:The equation which relates the number of loops in the standing wave and length of the standing wave of a string with two fixed ends can be used here. By using the de Broglie wavelength equation, the relationship between momentum of particle and length can be determined. Then substituting the momentum equation in terms of length in the energy equation, we can find out the relation between energy and length.
Formula used:
\[L=\dfrac{n\lambda }{2}\]
\[p=\dfrac{h}{\lambda }\]
\[E=\dfrac{{{p}^{2}}}{2m}\]

Complete answer:
Given that a particle is restricted to move along the x-axis between \[x=0\] and \[x=a\].
If there are \[n\] loops in the standing wave, we have, \[L=\dfrac{n\lambda }{2}\]
Where,
\[L\] is the length of the standing wave of a string with two fixed ends.
\[\lambda \] is the wavelength
\[n\] is the number of loops.
Given that, \[L=a\]. Then,
\[a=\dfrac{n\lambda }{2}\]
Then,
\[\lambda =\dfrac{2a}{n}\] -------- 1
From de Broglie wavelength formula,
\[p=\dfrac{h}{\lambda }\] --------- 2
Where,
\[p\] is the momentum of the particle
\[h\] is the Planck's constant
Substitute 2 in equation 1. We get,
\[p=\dfrac{nh}{2a}\] -------- 3
Given that,
Energy, \[E=\dfrac{{{p}^{2}}}{2m}\]-------- 4
Substitute 3 in 4, we get,
\[E=\dfrac{{{n}^{2}}{{h}^{2}}}{8{{a}^{2}}m}\]
From the above equation \[E\propto {{a}^{-2}}\]
The allowed energy for the particle for a particular value of n is proportional to \[{{a}^{-2}}\]

So, the correct answer is “Option A”.

Note:
A standing wave is a pattern obtained from the interference of two or more waves traveling in the same medium. Standing waves are characterized by nodes and anti nodes. Positions along the medium which are standing still are known as nodes. Antinodes are positions along the medium where the particles oscillate about their equilibrium position with maximum amplitude.