CBSE Class 12 Physics 2025: question 27

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Step-by-step solution

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1. Postulate 1: the centripetal force is provided by the Coulomb attraction, and the electron moves in certain stable (stationary) orbits without radiating: $\frac{m{v}^2}{r}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$ ...(1)

   ½ mark

2. Postulate 2 (quantisation of angular momentum): $L=mvr=\frac{nh}{2\pi }$, n = 1, 2, 3, ... ...(2)

   ½ mark

3. Postulate 3 (frequency condition): $h\nu =E_i-E_f$ when the electron jumps from a higher orbit to a lower one.

   ½ mark

4. (a) From (2), $v=\frac{nh}{2\pi mr}$. Substituting into (1), $m{v}^{2}r=\frac{e^2}{4\pi {\epsilon }_0}$ gives $m\frac{n^2{h}^2}{4{\pi }^2{m}^2{r}^2}r=\frac{e^2}{4\pi {\epsilon }_0}$.

   ½ mark

5. $r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2}$. Hence $r_n\propto n^2$.

   ½ mark

6. (b) KE $=\frac{1}{2}m{v}^2=\frac{e^2}{8\pi {\epsilon }_{0}r}$ and PE $=-\frac{e^2}{4\pi {\epsilon }_{0}r}$, so $E_n=-\frac{e^2}{8\pi {\epsilon }_0{r}_n}$. Substituting $r_n$: $E_n=-\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$. Hence $E_n\propto \frac{1}{n^2}$.

   ½ mark

Answer: r_n = n²h²ε₀/(πme²) ∝ n²; E_n = −me⁴/(8ε₀²h²n²) ∝ 1/n².

Revise the chapter

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