JEE Advanced · Physics · 27. Atomic Physics
Paragraph:
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.
Question:
A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the \(n\)th level ( \(n=0\) is not allowed) is
- A \(\frac{1}{n^2}\left(\frac{h^2}{8 \pi^2 I}\right)\)
- B \(\frac{1}{n}\left(\frac{h^2}{8 \pi^2 I}\right)\)
- C \(n\left(\frac{h^2}{8 \pi^2 I}\right)\)
- D \(n^2\left(\frac{h^2}{8 \pi^2 I}\right)\)
Answer & Solution
Correct Answer
(D) \(n^2\left(\frac{h^2}{8 \pi^2 I}\right)\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
L=I \omega & =\frac{n h}{2 \pi} \therefore \omega=\frac{n h}{2 \pi I} \\
K & =\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{n h}{2 \pi I}\right)^2 \\
& =\frac{n^2 h^2}{8 \pi^2 I}
\end{aligned}
\)
\(\therefore\) The correct answer is (d).
\begin{aligned}
L=I \omega & =\frac{n h}{2 \pi} \therefore \omega=\frac{n h}{2 \pi I} \\
K & =\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{n h}{2 \pi I}\right)^2 \\
& =\frac{n^2 h^2}{8 \pi^2 I}
\end{aligned}
\)
\(\therefore\) The correct answer is (d).
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Physics
- Statement I The formula connecting \(u, v\) and \(f\) for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.
Statement II Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.JEE Advanced 2007 Hard - Paragraph : In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, stand for dimensions of electric and magnetic fields respectively, while and stand for dimensions of the permittivity and permeability of free space respectively. are dimensions of length and time respectively. All the quantities are given in units
Question : The relation between isJEE Advanced 2018 Easy - A projectile is thrown from a point \(\mathrm{O}\) on the ground at an angle \(45^{\circ}\) from the vertical and with a speed \(5 \sqrt{2} \mathrm{~m} / \mathrm{s}\). The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, \(0.5 \mathrm{~s}\) after the splitting. The other part, \(t\) seconds after the splitting, falls to the ground at a distance \(x\) meters from the point O. The acceleration due to gravity \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).
The value of is ________.JEE Advanced 2021 Easy - A circular wire loop of radius \(R\) is placed in the \(x-y\) plane centered at the origin \(O\). A square loop of side \(a(a< < R)\) having two turns is placed with its centre at \(z=\sqrt{3} R\) along the axis of the circular wire loop, as shown in figure.
The plane of the square loop makes an angle of \(45^{\circ}\) with respect to the \(z\)-axis. If the mutual inductance between the loops is given by \(\frac{\mu_{0} a^{2}}{2^{p / 2} R}\), then the value of \(p\) isJEE Advanced 2012 Hard - An infinitely long solid cylinder of radius \(R\) has a uniform volume charge density \(\rho\). It has a spherical cavity of radius \(R / 2\) with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point \(P\), which is at a distance \(2 R\) from the axis of the cylinder, is given by the expression \(\frac{23 \rho R}{16 K \varepsilon_{0}}\). The value of \(k\) is
JEE Advanced 2012 Medium - In a Young's double slit experiment, a combination of two glass wedges \(A\) and \(B\), having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is \(d=2 \mathrm{~mm}\) and the shortest distance between the slits and the screen is \(D=2 \mathrm{~m}\). Thickness of the combination of the wedges is \(t=12 \mu \mathrm{~m}\). The value of \(l\) as shown in the figure is 1 mm . Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm ) with respect to O by ______
JEE Advanced 2025 Hard
More PYQs from JEE Advanced
- For , the number of real roots of the equation isJEE Advanced 2021 Easy
- Let be a differentiable function with . If satisfies the differential equation , then the value of isJEE Advanced 2018 Easy
- Let be the number of red and black balls, respectively, in box I. Let be the number of red and black balls, respectively, in box II.
A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is , then the correct options(s) with the possible values of is(are)JEE Advanced 2015 Medium - The compound(s) which react(s) with to give boron nitride is(are)JEE Advanced 2022 Medium
- A particle \(P\) starts from the point \(z_0=1+2 i\), where \(i=\sqrt{-1}\). It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point \(z_1\). From \(z_1\) the particle moves \(\sqrt{2}\) units in the direction of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and then it moves through an angle \(\frac{\pi}{2}\) in anti-clockwise direction on a circle with centre at origin, to reach a point \(z_2\). The point \(z_2\) is given byJEE Advanced 2008 Medium
- As per IUPAC nomenclature, the name of the complex \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{Cl}_{3}\) is :JEE Advanced 2012 Easy