CBSE Class 12 Physics 2025: question 20

Hints

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Step-by-step solution

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1. Time period of revolution in the $n^{th}$ orbit: $T=\frac{2\pi r_n}{v_n}$ ...(1).

   ½ mark

2. Bohr's quantisation condition: $m{v}_n{r}_n=\frac{nh}{2\pi }\Rightarrow v_n=\frac{nh}{2\pi m{r}_n}$ ...(2). Substituting in (1): $T=\frac{4{\pi }^{2}m{r}_n^2}{nh}$.

   ½ mark

3. Radius of the $n^{th}$ orbit: $r_n=\frac{{\epsilon }_0{n}^2{h}^2}{\pi m{e}^2}$, so $r_n\propto n^2$.

   ½ mark

4. Hence $T=\frac{4{\pi }^{2}m}{nh}\cdot \frac{{\epsilon }_0^2{n}^4{h}^4}{{\pi }^2{m}^2{e}^4}=\frac{4{\epsilon }_0^2{n}^3{h}^3}{m{e}^4}$, so $T\propto n^3$. (Shortcut: $r\propto n^2$ and $v\propto \frac{1}{n}$ give $T=\frac{2\pi r}{v}\propto n^3$.)

   ½ mark

Answer: T = 4ε₀²n³h³/(me⁴), so T ∝ n³

Revise the chapter

• Chapter: Atoms (notes, formulas, where marks are lost and every past question)
• Topic: Bohr Model of the Hydrogen Atom