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) the 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
\(\lambda\) = wavelength of radiation
Substituting the values in the given expression of (v): v = \(\frac{3.0\times10^8\,m/s}{616\times10^{-9}m}\)
= 4.87 × 10⁸ × 10⁹ × 10³
= 4.87 × 10¹⁴
Frequency of emission = 4.87×10 ¹⁴ m

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(b) Velocity of radiation, (c) = 3.0×10⁸ ms 
Distance travelled by this radiation in 30 s = (3.0×10 ⁸ ms) (30s) = 9.0 × 10⁹ m

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(c) Energy of quantum (E) = (6.626 × 10³⁴Js) (4.87×10¹⁴ s)
The energy of quantum (E) = 32.27×10^-20

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(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 = \(\frac{2J}{32.27\times 10^{-20}J}\) = 6.19×10¹⁸
= 6.2×10¹⁸