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Question:

In a hydrogen-like atom, an electron makes a transition from an energy level with the quantum number n to another with the quantum number (n-1). If n>>1, the frequency of radiation emitted is approximately proportional to:

1n3/2

1n3

1n

1n2

Solution:

ΔE = hμ
The energy of an electron in a hydrogen-like atom is given by:
E = -13.6 Z²/n² eV
where Z is the atomic number and n is the principal quantum number.
For a transition from n to (n-1), the change in energy is:
ΔE = E(n-1) - E(n) = -13.6 Z²[(1/(n-1)²) - (1/n²)]
Since n>>1, we can approximate (n-1)² ≈ n²
ΔE ≈ -13.6 Z² [1/n² - 1/n²] = -13.6 Z² [n² - (n-1)²] / (n²(n-1)²) ≈ -13.6 Z² (2n -1) / n⁴
For large n, ΔE ≈ -13.6 Z² (2n) / n⁴ = -27.2 Z²/n³
Since ΔE = hμ, the frequency μ is proportional to ΔE:
μ ∝ 1/n³
Therefore, the frequency of radiation emitted is approximately proportional to 1/n³.