The energy $$E$$ of a photon is related to its wavelength $$\lambda$$ by the formula:

$$E=\frac{hc}{\lambda}$$

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:

$$\mathrm{Energy~absorbed}=E_1-E_2=\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)$$

Substituting the given values for $$h$$ and $$c$$, we get:

$$\mathrm{Energy~absorbed}=6.6\times10^{-34}~\mathrm{J\cdot s}\cdot3\times10^8~\mathrm{m/s}\cdot\left(\frac{1}{500\times10^{-9}~\mathrm{m}}-\frac{1}{600\times10^{-9}~\mathrm{m}}\right)$$

Simplifying this expression, we get:

$$\mathrm{Energy~absorbed}=6.6\times10^{-20}~\mathrm{J}$$

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:

$$\mathrm{Energy~absorbed~in~eV}=\frac{6.6\times10^{-20}~\mathrm{J}}{1.6\times10^{-19}~\mathrm{C/eV}}=0.4125~\mathrm{eV}$$

Finally, we can express the net energy absorbed in terms of the given value of $$n$$, as follows:

$$n\times10^{-4}~\mathrm{eV}=0.4125~\mathrm{eV}$$

Solving for $$n$$, we get:

$$n=\frac{0.4125}{10^{-4}}=4125$$

Therefore, the value of $$n$$ is $$\boxed{4125}$$.