Question:medium

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 x 10-34 js and c = 3 x 108 m/s)

Updated On: Feb 26, 2026
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Correct Answer: 4125

Solution and Explanation

To solve the problem, we need to determine the net energy absorbed by the atom. The atom first absorbs a photon of wavelength 500 nm and emits a photon of wavelength 600 nm. We use the formula for the energy of a photon: E = hc/λ, where h is Planck's constant and c is the speed of light.
First, calculate the energy absorbed by the photon of wavelength 500 nm:
Eabsorb = hc/λ1 = (6.6 × 10-34 J·s)(3 × 108 m/s) / (500 × 10-9 m) = 3.96 × 10-19 J
Next, calculate the energy emitted by the photon of wavelength 600 nm:
Eemit = hc/λ2 = (6.6 × 10-34 J·s)(3 × 108 m/s) / (600 × 10-9 m) = 3.3 × 10-19 J
Then, find the net energy absorbed by the atom:
ΔE = Eabsorb − Eemit = (3.96 × 10-19 J) − (3.3 × 10-19 J) = 0.66 × 10-19 J
Convert the net energy from joules to electron volts using the conversion factor: 1 eV = 1.6 × 10-19 J.
ΔE = 0.66 × 10-19 J / (1.6 × 10-19 J/eV) = 0.4125 eV
The net energy absorbed is expressed as n × 10-4 eV. Hence, n = 4125.
Verify that n = 4125 is within the expected range [4125, 4125], which it is.
Therefore, the value of n is 4125.
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