Question

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy

Answer 1

Wavelength of radiation emitted = 616 nm = 616 × ${10}^{–9}$ m (Given) (a) Frequency of emission (v)
$v=\frac{c}{\lambda }$
Where, c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of (v):
$v=\frac{3.0\times {10}^{8}m/s}{616\times {10}^{-9m}}=4.87\times {10}^8\times {10}^9\times {10}^{-3}{s}^{-1}v=4.87\times {10}^{14}{s}^{-1}$
Frequency of emission (ν) = 4.87 × ${10}^{14}{s}^{–1}$
(b) Velocity of radiation, (c) = 3.0 × ${10}^{8}m{s}^{–1}$
Distance travelled by this radiation in 30 s
$=(3.0\times {10}^{8}m{s}^{-1})\left(30s\right)=9.0\times {10}^{9}m$
(c) Energy of quantum (E) = hν
$(6.626\times {10}^{-34}Js)(4.87\times {10}^{14}{s}^{-1}$
Energy of quantum (E) = 32.27 × ${10}^{–20}$ J
(d) Energy of one photon (quantum) = 32.27 × ${10}^{–20}$ J
Therefore, 32.27 × ${10}^{–20}$ J of energy is present in 1 quantum. Number of quanta in 2 J of energy
$\frac{2J}{32.27\times {10}^{-20}J}=6.19\times {10}^{18}=6.2\times {10}^{18}$