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

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Last updated date: 20th Jun 2024
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Answer
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Hint: According to planck's theory, the energy of the radiation is expressed as follows,
$\text{ E = h }\nu \text{ = }\dfrac{\text{hC }}{\lambda }\text{ }$
Where, h is Planck's constant, C is the speed of light, $\lambda $ is wavelength of radiation. The energy of the radiation always about the law of conservation of energy. According to which the total energy of the photon absorbs by the particle is equal to the energy emitted by the photon. Thus,
$\text{ }{{\text{E}}_{\text{Total}}}\text{ = }{{\text{E}}_{\text{1}}}\text{ + }{{\text{E}}_{\text{2 }}}$
Where $\text{ }{{\text{E}}_{\text{1}}}\text{ }$ is the energy of the first radiation and $\text{ }{{\text{E}}_{\text{2}}}\text{ }$ is the energy emitted by the second photon

Complete answer:
According to Planck's quantum theory, the energy of light radiation is directly proportional to the frequency of radiation. The frequency is denoted by the $\text{ }\nu \text{ }$. The proportionality constant relates the frequency and the energy by the planck's constant ‘h’. The energy of the light radiation is given as follows,
$\text{ E = h }\nu \text{ }$
The energy can be also written in terms of wavelength is,
$\text{ E = h }\nu \text{ = }\dfrac{\text{hC }}{\lambda }\text{ }$
Where, h is Planck's constant, C is the speed of light, $\lambda $ is wavelength of radiation.
Let's consider a gas molecule is a strike with a photon of radiation. Then, the law of conservation of energy of the radiation absorbed is equal to the energy emitted by the two photons. This is given as follows,
$\text{ }{{\text{E}}_{\text{Total}}}\text{ = }{{\text{E}}_{\text{1}}}\text{ + }{{\text{E}}_{\text{2 }}}$ (1)
Where $\text{ }{{\text{E}}_{\text{1}}}\text{ }$ is the energy of the first radiation and $\text{ }{{\text{E}}_{\text{2}}}\text{ }$is the energy emitted by the second photon. Using the planck's quantum theory, the above relation (1) is written as,
$\text{ }\dfrac{\text{hC }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{Total}}}}\text{ = }\dfrac{\text{hC }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}}}\text{ + }\dfrac{\text{hC }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{2}}}}\text{ }$
On further simplifying we have,
$\text{ }\dfrac{\text{1 }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{Total}}}}\text{ = }\dfrac{\text{1 }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{1}}}}\text{ + }\dfrac{\text{1 }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{2}}}}\text{ }$
Let's consider a gas absorbs a photon of the wavelength 355 nm and emits the radiation of the wavelength 680 nm. Then the wavelength of the other radiation emitted by the photon is calculated as follows,
$\begin{align}
& \text{ }\dfrac{\text{1 }}{355}\text{ = }\dfrac{\text{1 }}{680}\text{ + }\dfrac{\text{1 }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{2}}}}\text{ } \\
& \Rightarrow \dfrac{\text{1 }}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{2}}}}\text{ = }\dfrac{\text{1 }}{355}\text{ }-\dfrac{\text{1 }}{680} \\
& \therefore {{\text{ }\!\!\lambda\!\!\text{ }}_{\text{2}}}\text{ = 742}\text{.76 }\simeq \text{ 743 nm } \\
\end{align}$
Therefore, the second wavelength radiated by the photon is equal to $\text{ 743 nm }$ .

Hence, (C) is the correct option.

Note: Note that such questions always test your presence of mind. We should always remember that the total energy of a system is always conserved.it is neither created nor destroyed but it can be transformed into one from to another. Let suppose a photon emits more ‘n’ number of radiation then the total energy of the system can be written as,
$\text{ }{{\text{E}}_{\text{Total }}}\text{= }\sum\limits_{\text{i}}^{\text{n}}{\dfrac{\text{hC}}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{i}}}}}\text{ }$ .