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$E=\dfrac{hc}{\lambda }$,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: $E=\dfrac{hc}{\lambda }$
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,
$\dfrac{hc}{{{\lambda }_{T}}}=\dfrac{hc}{{{\lambda }_{1}}}+\dfrac{hc}{{{\lambda }_{2}}}$
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,
$\dfrac{1}{{{\lambda }_{T}}}=\dfrac{1}{{{\lambda }_{1}}}+\dfrac{1}{{{\lambda }_{2}}}$
Consider this equation as equation (3),
Now by substituting the given values in equation (3) we get,
$\dfrac{1}{355}=\dfrac{1}{680}+\dfrac{1}{{{\lambda }_{2}}}$

& \Rightarrow \dfrac{1}{{{\lambda }_{2}}}=\dfrac{1}{355}-\dfrac{1}{680} \\\ & \Rightarrow \dfrac{1}{{{\lambda }_{2}}}=\dfrac{680-355}{355\times 680} \\\ & \Rightarrow {{\lambda }_{2}}=743nm \\\ \end{aligned}$$ **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.