Question

A photon of $300\,nm$ is absorbed by gas and then re-emits two photons. One re-emitted photon has wavelength $496\,nm$ , the wavelength of the second re-emitted photon is:
(A) $759$
(B) $857$
(C) $957$
(D) $657$

Answer 1

Hint : In this question, we have to find the wavelength of the second re-emitted photon. To find wavelength we are using the de Broglie equation to find the total energy than using that formula we calculate wavelength.

Complete Step By Step Answer:
The de Broglie equation relates a moving particle’s wavelength with its momentum. The de Broglie wavelength is the wavelength, $\lambda$ associated with massive particles and is related to its momentum $p$ through the Planck constant $h$ .
$\lambda =\dfrac{h}{p}$
In other words, you can say that matter also behaves like waves; this is what is called the de Broglie hypothesis.
$E=\dfrac{hc}{\lambda }$
We know that,
The wavelength of the First photon = ${{\lambda }_{1}}$
The wavelength of the Second photon = ${{\lambda }_{2}}$
$E(total)={{E}_{1}}+{{E}_{2}}$
We can write this equation as,
$\dfrac{hc}{\lambda }=\dfrac{hc}{{{\lambda }_{1}}}+\dfrac{hc}{{{\lambda }_{2}}}$
Now we substitute the values in the given equation,
$\dfrac{1}{300}=\dfrac{1}{496}+\dfrac{1}{{{\lambda }_{2}}}$
After solving this equation we get the value of the wavelength of the second photon,
${{\lambda }_{2}}=759\,nm$
So, option A is the correct answer.

Note :
To solve this question we know photons and the de Broglie equation. De Broglie’s hypothesis of matter waves postulates that any particle of matter that has linear momentum is also a wave. The wavelength of a matter-wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The speed of the matter waves is the speed of the particle. After understanding the concept of de Broglie we can easily solve this type of question.