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The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionization energy of H = 2.18 × 10⁻¹⁸ J atom⁻¹ and h = 6.625 × 10⁻³⁴ J s):

1. 1.54 × 10¹⁵ s⁻¹
2. 1.03 × 10¹⁵ s⁻¹
3. 3.08 × 10¹⁵ s⁻¹
4. 2.00 × 10¹⁵ s⁻¹

Correct answer: 1.54 × 10¹⁵ s⁻¹

Solution

The energy difference between n=4 and n=1 is calculated using the formula ΔE = -2.18 × 10⁻¹⁸ × (1/n₁² - 1/n₂²). Substituting n₁=1 and n₂=4, ΔE = 2.04375 × 10⁻¹⁸ J. Using E = hν, ν = ΔE/h = (2.04375 × 10⁻¹⁸)/(6.625 × 10⁻³⁴) = 1.54 × 10¹⁵ s⁻¹.

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