If the mass of the particle is and the energy of the particle in its ground state is closest 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. If the mass of the particle is $m = 1.0 \times 10^{- 30} kg$ and $a = 6.6 nm$ the energy of the particle in its ground state is closest 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. If the mass of the particle is $m = 1.0 \times 10^{- 30} kg$ and $a = 6.6 nm$ the energy of the particle in its ground state is closest to [2009]

Smart Achievers · Accepted Answer

(2)

For ground state, $x = 0$

Given, $m = 1.0 \times 10^{- 30} kg,$ $p$

$∴$ $E = \frac{n^{2} h^{2}}{8 m a^{2}} = \frac{1^{2} \times {(6.6 \times 10^{- 34})}^{2}}{8 \times 1 \times 10^{- 30} \times {(6.6 \times 10^{- 9})}^{2}} J$

$= 8 meV$

Solution

1 0.8 meV

2 8 meV

3 80 meV

4 800 meV

Ans.
(2)

For ground state, $n = 1$

Given, $m = 1.0 \times 10^{- 30} kg,$ $a = 6.6 \times 10^{- 9} m$

$∴$ $E = \frac{n^{2} h^{2}}{8 m a^{2}} = \frac{1^{2} \times {(6.6 \times 10^{- 34})}^{2}}{8 \times 1 \times 10^{- 30} \times {(6.6 \times 10^{- 9})}^{2}} J$

$= 8 meV$

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