Modern Physics 1 Question 31
28. In a $CO$ molecule, the distance between $C$ (mass $=12 amu$ ) and $O$ $($ mass $=16 amu)$, where $1 amu=\frac{5}{3} \times 10^{-27} kg$, is close to
(a) $2.4 \times 10^{-10} m$
(b) $1.9 \times 10^{-10} m$
(c) $1.3 \times 10^{-10} m$
(d) $4.4 \times 10^{-11} m$
(2010)
Passage 2
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 $E=\frac{p^{2}}{2 m}$. 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 $h=6.6 \times 10^{-34} Js$ and $e=1.6 \times 10^{-19} C$ ]
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Solution:
- $I=\mu r^{2}$ ( where, $\mu=$ reduced mass)
$$ \mu=\frac{m _1 m _2}{m _1+m _2}=\frac{48}{7} amu=11.43 \times 10^{-27} kg $$
Substituting in $I=\mu r^{2}$ we get,
$$ \begin{aligned} r & =\sqrt{\frac{I}{\mu}}=\sqrt{\frac{1.87 \times 10^{-46}}{11.43 \times 10^{-27}}} \\ & =1.28 \times 10^{-10} m \end{aligned} $$
