Question

A photon of $300nm$ is absorbed by a gas which then re-emits two photons. One re-emitted photon has wavelength $496nm$. Calculate the energy associated with the other photon re-emitted out.

• A. $1.911\times {10}^{-19}J$
• B. $1.04\times {10}^{-19}J$
• C. $3.3125\times {10}^{-19}J$
• D. $2.625\times {10}^{-19}J$

Answer 1

The correct option is D $2.625\times {10}^{-19}J$

$\lambda$ of the photon absorbed $=300$ nm

${\lambda }_1$ of $I$ photon re-emitted out $=496$ nm

Let ${\lambda }_2$ of $II$ photon re-emitted out $={\lambda }_{II}$

Then, total energy absorbed= Total energy re-emitted out

$\therefore$ $\frac{hc}{300\times {10}^{-9}}=\frac{hc}{496\times {10}^{-9}}+\frac{hc}{{\lambda }_2}$

$\therefore {\lambda }_2=759.18$ nm

Also, Energy re-emitted in the form of $II$ photon

$E_{Re}=\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{759.18\times {10}^{-9}}$

$=2.618\times {10}^{-19}J$