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?

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Answer 1 · 2017-05-22T09:47:52

iii)

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

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