A diatomic molecule is made of two masses m_1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The rotational kinetic energy $(RKE)$ of a system of two masses $m_1$ and $m_2$ separated by distance $r$ rotating about their center of mass is g

Vedclass · Accepted Answer

$\frac{(m_1 + m_2) n^2 h^2}{8 \pi^2 m_1 m_2 r^2}$

Vedclass · Answer

(A) The rotational kinetic energy $(RKE)$ of a system of two masses $m_1$ and $m_2$ separated by distance $r$ rotating about their center of mass is given by $RKE = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.The reduced mass is $\mu = \frac{m_1 m_2}{m_1 + m_2}$.The moment of inertia about the center of mass is $I = \mu r^2 = \frac{m_1 m_2}{m_1 + m_2} r^2$.According to Bohr's quantization rule,the angular momentum is $L = \frac{nh}{2\pi}$.Substituting these into the $RKE$ formula:$RKE = \frac{(nh/2\pi)^2}{2I} = \frac{n^2 h^2}{4\pi^2 \cdot 2 \cdot (\frac{m_1 m_2}{m_1 + m_2} r^2)} = \frac{n^2 h^2 (m_1 + m_2)}{8\pi^2 m_1 m_2 r^2}$.Note: The provided options in the prompt appear to have typographical errors regarding the $\pi$ factor. Based on standard physics,the correct expression is $\frac{(m_1 + m_2) n^2 h^2}{8 \pi^2 m_1 m_2 r^2}$.

$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:

Similar Questions

If in a hydrogen atom,the radius of the ${n^{th}}$ Bohr orbit is ${r_n}$ and the frequency of revolution of the electron in the ${n^{th}}$ orbit is ${f_n}$,choose the correct option.

When an electron is excited from its $4^{\text{th}}$ orbit to its $5^{\text{th}}$ stationary orbit,the change in the angular momentum of the electron is approximately (Planck's constant $= h = 6.63 \times 10^{-34} \text{ J s}$)

If the difference between $(n + 1)^{th}$ Bohr radius and $n^{th}$ Bohr radius is equal to the $(n - 1)^{th}$ Bohr radius, then the value of $n$ is:

The energy of $He^{+}$ ion in its first excited state is $......eV$ (The ground state energy for the Hydrogen atom is $-13.6\,eV$).

An electron in Bohr's hydrogen atom has an energy of $-3.4 \ eV$. The angular momentum of the electron is

Vedclass Test Series

Exam Paper Generator

Online Exam Module