Question

A gas absorbs a photon of 355 nm and emits two wavelengths. If one of the emission is at 680 nm, the other is at:
(A) 518 nm
(B) 1035 nm
(C) 325 nm
(D) 743 nm

Answer 1

Hint: A photon is the smallest discrete amount of electromagnetic radiation. It is a basic unit of all light. Here, we have to find emission of other wavelengths. And wavelength is related to energy by equation$E={\displaystyle \frac{hc}{\lambda }}$,where h is Planck’s constant, c is velocity of light , $\lambda$ is wavelength, E is energy.


Complete step by step solution:
- From the law of conservation of energy, that is energy of absorbed photons must be equal to combined energy of two emitted photons.
$E_T=E_1+E{}_2$
Consider this equation as equation (1),
Where, $E_T$=total energy of photon, $E_1$=energy of first emitted photon, $E_2$=energy of second emitted photon.
- Now, energy E and wavelength of a photon are related by the equation: $E={\displaystyle \frac{hc}{\lambda }}$
Consider this equation as equation (2),
Where h is Planck’s constant, c is velocity of light.
-Now by inserting values from equation (2) to equation (1) we get,
$${\displaystyle \frac{hc}{{\lambda }_T}}={\displaystyle \frac{hc}{{\lambda }_1}}+{\displaystyle \frac{hc}{{\lambda }_2}}$$
Where, ${\lambda }_T$=total wavelength, ${\lambda }_1$=first wavelength, ${\lambda }_2$=second wavelength
In this equation, the values of hc cancels out and we get the formula,
$${\displaystyle \frac{1}{{\lambda }_T}}={\displaystyle \frac{1}{{\lambda }_1}}+{\displaystyle \frac{1}{{\lambda }_2}}$$
Consider this equation as equation (3),
Now by substituting the given values in equation (3) we get,
$${\displaystyle \frac{1}{355}}={\displaystyle \frac{1}{680}}+{\displaystyle \frac{1}{{\lambda }_2}}$$
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{{\lambda }_2}}={\displaystyle \frac{1}{355}}-{\displaystyle \frac{1}{680}}\\ & \Rightarrow {\displaystyle \frac{1}{{\lambda }_2}}={\displaystyle \frac{680-355}{355\times 680}}\\ & \Rightarrow {\lambda }_2=743nm\end{array}$$

So, the correct answer is “Option D”.

Additional Information:
- It is found that Photons are always in motion and in vacuum, these travel at a constant speed of $2.998\times {10}^{8}m/s$. This is commonly called the speed of light, which is denoted by the letter c.
- Energy of a photon is also found to be dependent on its frequency (that is how fast the electric and magnetic field wiggle), the higher the frequency, the more energy the photon has.
- The law of conservation of energy can be seen in everyday examples like-this energy can be used to rotate the turbine of a generator to produce electricity. And in this process the potential energy of water can be tuned into kinetic energy which can on further process will become electric energy.

Note: -One should not get confused in the relation of wavelength and frequency with energy. Wavelength is denoted by symbol $\lambda$and frequency by symbol $\nu$ , and as wavelength increases energy decreases and as frequency increases energy also increases.
- We should not forget to write the unit after solving any question. Here, we can see that the unit of wavelength is nm.