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JEE Advance - Physics (2009 - Paper 1 Offline - No. 15)

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 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 $$E = {{{p^2}} \over {2m}}$$. 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 three questions for a particle moving in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

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 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 $$E = {{{p^2}} \over {2m}}$$. 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 three questions for a particle moving in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

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 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 $$E = {{{p^2}} \over {2m}}$$. 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 three questions for a particle moving in the line x = 0 to x = a. Take $$h = 6.6 \times {10^{ - 34}}$$ J-s and $$e = 1.6 \times {10^{ - 19}}$$ C.

The speed of the particle, that can take discrete values, is proportional to
$${n^{ - 3/2}}$$
$${n^{ - 1}}$$
$${n^{1/2}}$$
$$n$$

Explanation

To find the relationship between the speed of the particle and the quantum number $$n$$, we need to follow a few steps using the given information and fundamental quantum mechanical principles.

Firstly, for a particle moving in a one-dimensional box of length $$a$$, the allowed wavelengths $$\lambda$$ of standing waves can be given by:

$$ \lambda = \frac{2a}{n} $$

for $ n = 1, 2, 3, \ldots $.

The de Broglie relation links the wavelength $$\lambda$$ of a particle to its momentum $$p$$:

$$ \lambda = \frac{h}{p} $$

From this, we can express the momentum $$p$$ in terms of the quantum number $$n$$:

$$ \frac{h}{p} = \frac{2a}{n} $$

Solving for $$p$$, we get:

$$ p = \frac{nh}{2a} $$

The kinetic energy $$E$$ of the particle is related to its momentum $$p$$ by the equation:

$$ E = \frac{p^2}{2m} $$

Substituting $$ p = \frac{nh}{2a} $$ into the energy expression, we get:

$$ E = \frac{(nh/2a)^2}{2m} = \frac{n^2 h^2}{8ma^2} $$

The kinetic energy can also be written in terms of the speed $$v$$ of the particle:

$$ E = \frac{1}{2} m v^2 $$

Equating the two expressions for energy, we have:

$$ \frac{1}{2} m v^2 = \frac{n^2 h^2}{8ma^2} $$

Solving for the speed $$v$$, we get:

$$ v^2 = \frac{n^2 h^2}{4m^2 a^2} $$

Therefore, the speed $$v$$ of the particle is:

$$ v = \frac{nh}{2ma} $$

From the above expression, we see that the speed $$v$$ of the particle is directly proportional to $$n$$:

$$ v \propto n $$

Therefore, the speed of the particle is proportional to:

Option D: $$ n $$

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