The allowed energy for the particle for a particular value of is proportional to [2009]

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 $E = \frac{p^{2}}{2 m} .$ Thus, the energy of the particle can be denoted by a quantum number $' n'$ taking values $1, 2, 3, \dots$ ( $n = 1$ , 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 $e = 1.6 \times 10^{- 19} C$ .

Q. The allowed energy for the particle for a particular value of $n$ is proportional to [2009]

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 $e = 1.6 \times 10^{- 19} C$ .

Q. The allowed energy for the particle for a particular value of $n$ is proportional to [2009]

Smart Achievers · Accepted Answer

(1)

Energy, $E = \frac{h^{2}}{2 m λ^{2}}$ $(∵ E = \frac{P^{2}}{2 m} and P = \frac{h}{λ})$

The length in which the particle is restricted to move is

$a = \frac{n λ}{2}$ $\Rightarrow λ = \frac{2 a}{n}$

Putting this value of $λ$ , we get

$E = \frac{h^{2} n^{2}}{2 m \times 4 a^{2}} = \frac{n^{2} h^{2}}{8 m a^{2}}$

$∴$ $E \propto a^{- 2}$

Solution

1 $a^{- 2}$

2 $a^{- 3 / 2}$

3 $a^{- 1}$

4 $a^{2}$

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Solution

1 a-2

2 a-3/2

3 a-1

4 a2

Ans. (1) Energy, E=h22mλ2 (∵ E=P22m and P=hλ) The length in which the particle is restricted to move is a=nλ2⇒λ=2an Putting this value of λ, we get E=h2n22m×4a2=n2h28ma2 ∴ E∝a-2

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1 $a^{- 2}$

2 $a^{- 3 / 2}$

3 $a^{- 1}$

4 $a^{2}$