# A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at - i) 1035 nm ii) 325 nm iii) 743 nm iv) 518 nm?

May 22, 2017

iii)

#### Explanation:

From Law of Conservation of energy, energy of absorbed photon must be equal to combined energy of two emitted photons.

${E}_{T} = {E}_{1} + {E}_{2}$ .....(1)
where ${E}_{1}$ is Energy of first emitted photon emitted and ${E}_{2}$ is Energy of second emitted photon.

Energy $E$ and wavelength λ of a photon are related by the equation

E = (hc)/λ .....(2)
where $h$ is Planck's constant, $c$ is velocity of light.

Inserting values from (2) in (1) we get

(hc)/λ_T = (hc)/λ_1 + (hc)/λ_2
=>(1)/λ_T = (1)/λ_1 + (1)/λ_2 ......(3)

Substituting given values in (3) we get

1/355 = 1/680 + 1/λ_2
=> 1/λ_2=1/355 - 1/680
=> 1/λ_2=(680-355)/(355xx680)
=>λ_2 = 742.77 nm