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?

1 Answer
May 22, 2017

Answer:

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#