Question

A photon of $300nm$ is absorbed by gas and then re-emits two photons. One re-emitted photon has wavelength $496nm$, the wavelength of the second re-emitted photon is:

• A. 759
• B. 857
• C. 957
• D. 657

Answer 1

Answer: (A)

759

This symmetric loving nature of Nature gave rise to de Broglie relation.

The de Broglie equation relates a moving particle's wavelength with its momentum. The de Broglie wavelength is the wavelength, λ, associated with a massive particle and is related to its momentum, p, through the Planck constant, h

$\lambda$ $=$ $\frac{h}{p}$

In other words, you can say that matter also behaves like waves. This is what is called as de Broglie hypothesis.

$E=$ $\frac{hc}{\lambda }$

wavelength of $1^{st}$ photon = ${\lambda }_1$

wavelength of $2^{nd}$ photon = ${\lambda }_2$

E(total) = E1 + E2 = Emitted energy

$\frac{hc}{\lambda }$ = $\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}$

$\frac{1}{300}$ = $\frac{1}{496}+\frac{1}{{\lambda }_2}$

${\lambda }_2$ = $759nm$ = wavelength of second photon.

So, the corret option is $A$