An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 600 nm. The net energy absorbed by the atom in this process is n × 10⁻⁴ eV. The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take h = 6.6 × 10⁻³⁴ Js and c = 3 × 10⁸ m/s) Numerical answer type. Work it out, then open the answer.

Answer: 4125. hc = 6.6 × 10⁻³⁴ × 3 × 10⁸ = 1.98 × 10⁻²⁵ J m = 1237.5 eV nm Net energy = hc(1/500 - 1/600) = (1237.5)/(3000) eV = 0.4125 eV = 4125 × 10⁻⁴ eV So n = 4125.

An atom absorbs a photon of wavelength 500 nm and emits another… | JEE Main 2023 (13 Apr, Shift 2)

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Other / Mixed: Atoms & Bohr Links

An atom absorbs a photon of wavelength $500\text{ nm}$ and emits another photon of wavelength $600\text{ nm}$ . The net energy absorbed by the atom in this process is $n \times 10^{-4}\text{ eV}$ . The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take $h = 6.6 \times 10^{-34}\text{ Js}$ and $c = 3 \times 10^{8}\text{ m/s}$ )

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

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Answer: 4125

$hc = 6.6 \times 10^{-34} \times 3 \times 10^{8} = 1.98 \times 10^{-25}\text{ J m} = 1237.5\text{ eV nm}$
Net energy $= hc\left(\frac{1}{500} - \frac{1}{600}\right) = \frac{1237.5}{3000}\text{ eV}$
$= 0.4125\text{ eV} = 4125 \times 10^{-4}\text{ eV}$
So $n = 4125$ .

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