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**Hint:**Think about the law of conservation of energy along with the equation for the De Broglie wavelength. Modify these equations to arrive at the correct answer.

**Formula used:**

**Complete step by step answer:**

Due to the law of conservation of energy, we know that the energy of the photon absorbed and the energies of the photons emitted is going to be the same. We will modify the equation for the De Broglie wavelength to suit our needs and use it along with the law of conservation of energies.

Let us first assume that,

$E$ = energy of absorbed photon

${{E}_{1}}$ = energy of first emitted photon

${{E}_{2}}$ = energy of second emitted photon

$\lambda $ = wavelength of absorbed photon

${{\lambda }_{1}}$ = wavelength of first emitted photon

${{\lambda }_{2}}$ = wavelength of second emitted photon

We know from the law of conservation of energies that:

\[E={{E}_{1}}+{{E}_{2}}\]

The De Broglie’s equation for wavelength states that:

\[\lambda =\frac{h}{mv}\]

Where, $m$ is the mass of the photon and $v$ is its velocity, which is equal to the speed of light $c$.

We know that, for a photon,

\[E=m{{c}^{2}}\]

Thus, modifying this equation,

\[mc\text{ or }mv=\frac{E}{c}\]

Putting this value in De Broglie’s equation we get

\[E=\frac{hc}{\lambda }\]

Now, substituting this value of $E$ in the law of conservation of energy, we get:

\[\frac{hc}{\lambda }=\frac{hc}{{{\lambda }_{1}}}+\frac{hc}{{{\lambda }_{2}}}\]

We know the values, \[\lambda =300nm\] and \[{{\lambda }_{1}}=760nm\]. Solving for \[{{\lambda }_{2}}\]:

\[\frac{1}{300} = \frac{1}{760} + \frac{1}{{{\lambda }_{2}}}\]

\[\frac{1}{{{\lambda }_{2}}} = \frac{1}{300} - \frac{1}{760}\]

\[\frac{1}{{{\lambda }_{2}}} = \frac{46}{22800}\]

\[{{\lambda }_{2}} = 495.6nm\]

**Hence, comparing the answers with the given options, the correct answer is ‘(B) 496nm’.**

**Note:**While modifying the De Broglie’s equation do not get confused amongst the values of $v$ and $c$. The former refers to the velocity of any particle, but since we are talking about photons here, velocity of the particle will always be equal to the velocity of light.

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