Q55 / 60 · JEE Advanced 2009 (Paper 1) · Physics
Paragraph:
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 liner 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 $1,2,3, \ldots(n=1$, 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 $h=6.6 \times 10^{-34} \mathrm{Js}$ and $e=1.6 \times 10^{-19} \mathrm{C}$ ]Question:
The speed of the particle that can take discrete values is proportional to
- A$n^{-3 / 2}$
- B$n^{-1}$
- C$n^{1 / 2}$
- Correct answerD$n$
Solution
From Eq. (i)
Hence, the correct answer is option: 'D'.