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JEE Advance - Physics (2009 - Paper 1 Offline - No. 13)

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 allowed energy for the particle for a particular value of $$n$$ is proportional to
$${a^{ - 2}}$$
$${a^{ - 3/2}}$$
$${a^{ - 1}}$$
$${a^2}$$

Explanation

We have,

$$a = n\left( {{\lambda \over 2}} \right) \Rightarrow \lambda = {{2a} \over n}$$

IIT-JEE 2009 Paper 1 Offline Physics - Dual Nature of Radiation Question 6 English Explanation

From de Broglie relation, we have

$$\lambda = {h \over {mv}} = {h \over p}$$

Therefore, $${{2a} \over n} = {h \over p} \Rightarrow p = {{nh} \over {2a}}$$

Now, $$E = {{{p^2}} \over {2m}} = {{{n^2}{h^2}} \over {8m{a^2}}} \Rightarrow E \propto {a^{ - 2}}$$

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