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IIT JEE Advanced 2009 Paper 1

Section: Physics

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Question : 15 of 60

Marks: +1, -0

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 =

\frac{p}{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 ×

10^{-34}\,\text{J-s}

10^{-19}\,\text{C}

The speed of the particle, that can take discrete values, is proportional to

[JEE Adv 2009 P1]

$n^{-\frac{3}{2}}$
$n^{-1}$
$n^{\frac{1}{2}}$
n

Solution:

We have the velocity as

V =

\frac{P}{m}

However, the momentum is given by

P =

\frac{nh}{2a}

Therefore,

V =

\frac{nh}{2ma}

⇒ V α n

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