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 $$n \times 10^{-4}$$ eV. The value of $$n$$ is [Assume the atom to be stationary during the absorption and emission process] (Take $$h = 6.6 \times 10^{-34}$$ J s and $$c = 3 \times 10^8$$ m s$$^{-1}$$).

The atom absorbs a photon of wavelength 500 nm and emits a photon of wavelength 600 nm.

Energy absorbed: $$E_1 = \frac{hc}{\lambda_1} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}}$$

$$E_1 = \frac{19.8 \times 10^{-26}}{5 \times 10^{-7}} = 3.96 \times 10^{-19} \text{ J}$$

Energy emitted: $$E_2 = \frac{hc}{\lambda_2} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}}$$

$$E_2 = \frac{19.8 \times 10^{-26}}{6 \times 10^{-7}} = 3.30 \times 10^{-19} \text{ J}$$

Net energy absorbed:

$$\Delta E = E_1 - E_2 = (3.96 - 3.30) \times 10^{-19} = 0.66 \times 10^{-19} \text{ J}$$

Converting to eV: $$\Delta E = \frac{0.66 \times 10^{-19}}{1.6 \times 10^{-19}} = 0.4125 \text{ eV}$$

$$\Delta E = 4125 \times 10^{-4} \text{ eV}$$

Therefore $$n = \mathbf{4125}$$.