Question 2.47: 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
Wavelength of radiation emitted = 616 nm = 616 × 10-9 m (Given)
(a) Frequency of emission
Where,
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression
= 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
= 6.19 ×1018
= 6.2 ×10
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Answer
Wavelength of radiation emitted = 616 nm = 616 × 10-9 m (Given)
(a) Frequency of emission
Where,
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression
= 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
= 6.19 ×1018
= 6.2 ×10
18
