A hydrogen atom changes its state from n = 3 to n = 2. Due to recoil, the percentage change in the wavelength of emitted light is approximately 1 × 10⁻ⁿ. The value of n is _______. [Given Rhc = 13.6 eV, hc = 1242 eV nm, h = 6.6 × 10⁻³⁴ J s mass of the hydrogen atom = 1.6 × 10⁻²⁷ kg ] Numerical answer type. Work it out, then open the answer.

Answer: 7. Photon energy: E = 13.6(1/4 - 1/9) = 1.89 eV. Recoil energy = p²/2M = (E²)/(2Mc²), so (Δλ)/(λ) = (Δ E)/(E) = (E)/(2Mc²). Mc² = 1.6 × 10⁻²⁷ × 9 × 10¹⁶1.6 × 10⁻¹⁹ = 9 × 10⁸ eV. (Δλ)/(λ) = 1.891.8 × 10⁹ ≈ 1 × 10⁻⁹, i.e. ≈ 1 × 10⁻⁷ %. So n = 7.

A hydrogen atom changes its state from n = 3 to n = 2. Due to recoil… | JEE Main 2024 (4 Apr, Shift 1)

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Transitions, Photon Emission & Excitation

A hydrogen atom changes its state from $n = 3$ to $n = 2$ . Due to recoil, the percentage change in the wavelength of emitted light is approximately $1 \times 10^{-n}$ . The value of $n$ is _______. [Given Rhc $= 13.6\text{ eV},\ hc = 1242\text{ eV nm},\ h = 6.6 \times 10^{-34}\text{ J s}$ mass of the hydrogen atom $= 1.6 \times 10^{-27}\text{ kg}$ ]

Numerical answer type. Work it out, then open the answer.

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Answer: 7

Photon energy: $E = 13.6\left(\frac{1}{4} - \frac{1}{9}\right) = 1.89\text{ eV}$ .
Recoil energy $= \frac{p^2}{2M} = \frac{E^2}{2Mc^2}$ , so $\frac{\Delta\lambda}{\lambda} = \frac{\Delta E}{E} = \frac{E}{2Mc^2}$ .
$Mc^2 = \frac{1.6 \times 10^{-27} \times 9 \times 10^{16}}{1.6 \times 10^{-19}} = 9 \times 10^{8}\text{ eV}$ .
$\frac{\Delta\lambda}{\lambda} = \frac{1.89}{1.8 \times 10^{9}} \approx 1 \times 10^{-9}$ , i.e. $\approx 1 \times 10^{-7}\ \%$ .
So $n = 7$ .

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