Question 247: Dual Nature of Matter and Radiation

Question

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

Accepted Answer

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:
$$\mathrm{Energy~absorbed}=E_1-E_2=\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}ight)$$
Substituting the given values for $$h$$ and $$c$$, we get:
$$\mathrm{Energy~absorbed}=6.6	imes10^{-34}~\mathrm{J\cdot s}\cdot3	imes10^8~\mathrm{m/s}\cdot\left(\frac{1}{500	imes10^{-9}~\mathrm{m}}-\frac{1}{600	imes10^{-9}~\mathrm{m}}ight)$$
Simplifying this expression, we get:
$$\mathrm{Energy~absorbed}=6.6	imes10^{-20}~\mathrm{J}$$
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:
$$\mathrm{Energy~absorbed~in~eV}=\frac{6.6	imes10^{-20}~\mathrm{J}}{1.6	imes10^{-19}~\mathrm{C/eV}}=0.4125~\mathrm{eV}$$
Finally, we can express the net energy absorbed in terms of the given value of $$n$$, as follows:
$$n	imes10^{-4}~\mathrm{eV}=0.4125~\mathrm{eV}$$
Solving for $$n$$, we get:
$$n=\frac{0.4125}{10^{-4}}=4125$$
Therefore, the value of $$n$$ is $$\boxed{4125}$$.

Dual Nature of Matter and Radiation

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