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

The energy $E$ of a photon is related to its wavelength $\lambda$ by the formula: <br/><br/> $E=\frac{hc}{\lambda}$ <br/><br/> where $h$ is Planck's constant and $c$ is the speed of light. In this problem, we are given that an atom absorbs a photon of wavelength $\lambda_1=500~\mathrm{nm}$ and emits another photon of wavelength $\lambda_2=600~\mathrm{nm}$. We can use the formula above to calculate the energy absorbed by the atom: <br/><br/> $$\mathrm{Energy~absorbed}=E_1-E_2=\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$$ <br/><br/> Substituting the given values for $h$ and $c$, we get: <br/><br/> $$\mathrm{Energy~absorbed}=6.6\times10^{-34}~\mathrm{J\cdot s}\cdot3\times10^8~\mathrm{m/s}\cdot\left(\frac{1}{500\times10^{-9}~\mathrm{m}}-\frac{1}{600\times10^{-9}~\mathrm{m}}\right)$$ <br/><br/> Simplifying this expression, we get: <br/><br/> $\mathrm{Energy~absorbed}=6.6\times10^{-20}~\mathrm{J}$ <br/><br/> We need to express this energy in electron volts (eV), which is a more convenient unit for atomic and molecular energies. To do this, we can divide the energy in joules by the charge of an electron: <br/><br/> $$\mathrm{Energy~absorbed~in~eV}=\frac{6.6\times10^{-20}~\mathrm{J}}{1.6\times10^{-19}~\mathrm{C/eV}}=0.4125~\mathrm{eV}$$ <br/><br/> Finally, we can express the net energy absorbed in terms of the given value of $n$, as follows: <br/><br/> $n\times10^{-4}~\mathrm{eV}=0.4125~\mathrm{eV}$ <br/><br/> Solving for $n$, we get: <br/><br/> $n=\frac{0.4125}{10^{-4}}=4125$ <br/><br/> Therefore, the value of $n$ is $\boxed{4125}$.