CBSE Class 12 Physics 2023: question 30(b)

Hints

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Step-by-step solution

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1. (i) Postulate 1: An electron revolves round the nucleus only in certain stable orbits (stationary states). While in these orbits it does not radiate energy.

   ½ mark

2. Postulate 2: The allowed orbits are those for which the angular momentum is an integral multiple of $\frac{h}{2\pi }$, i.e. $L=mvr=\frac{nh}{2\pi }$, where $n=1,2,3,\dots$

   ½ mark

3. Postulate 3: When an electron jumps from a higher-energy orbit to a lower-energy orbit, it emits a photon whose energy equals the energy difference: $h\nu =E_i-E_f$.

   ½ mark

4. (ii) Using $L=\frac{nh}{2\pi }$ with $n=2$: $L=\frac{2h}{2\pi }=\frac{h}{\pi }$.

   ½ mark

5. Substituting: $L=\frac{6.63\times 10^{-34}}{3.14}$.

   ½ mark

6. $L\approx 2.1\times 10^{-34}{\text{kg m}}^2{\text{s}}^{-1}$

   ½ mark

Answer: (ii) L = h/π ≈ 2.1 × 10⁻³⁴ kg m² s⁻¹

Revise the chapter

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