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Question
Neon gas is generally used in the signboards. If it emits strongly at 616 nm, calculate
- the frequency of emission,
- distance traveled by this radiation in 30 s
- energy of quantum and
- number of quanta present if it produces 2 J of energy.
Solution
Wavelength of radiation emitted = 616 nm = 616 × 10–9 m (Given)
a) Frequency of emission (v)
`"v" = "c"/lambda`
Where,
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of (v):
`"v" = (3.0xx10^8" m/s")/(616xx10^(-9) " m")`
= 4.87 × 108 × 109 × 10–3 s-1
ν = 4.87 × 1014 s–1
Frequency of emission (ν) = 4.87 × 1014 s-1
(b) Velocity of radiation, (c) = 3.0 × 108 ms-1
Distance travelled by this radiation in 30 s
= (3.0 × 108 ms-1) (30 s)
= 9.0 × 109 m
(c) Energy of quantum (E) = hν
(6.626 × 10-34 Js) (4.87 × 1014 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
`= "2J"/(32.27 xx 10^(-20)" J")`
`= 6.19 xx 10^(18)`
= 6.2 ×1018
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