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

Explanation

Wavelength of radiation emitted = 616 nm` = 616 × 10^(-9) m` (Given)
(a) Frequency of emission (v)
`v = (c)./(lambda)`
c = velocity of radiation
`lambda =` wavelength of radiation
Substituting the values in the given expression of (v) :
`v = (3.0 xx 10^(8) m//s)/(616 xx 10^(-9) m)`
`= 4.87 xx 10^(8) xx 10^(9) xx 10^(-3) s^(-1)`
`v = 4.87 xx 10^(14) s^(-1)`
Frequency of emission `(v) = 4.87 xx 10^(14) s^(-1)`
(b) Velocity of radiation, `(c). = 3.0 × 10^(8) ms^(1)`
Distance travelled by this radiation in 30 s
`= (3.0 xx 10^(8)ms^(-1))(30s)`
`= 9.0 xx 10^(9) m`
(c). Energy of quantum `(E) = hν`
`(6.626 xx 10^(-34)Js)(4.87 xx 10^(14)s^(-1))`
Energy of quantum `(E) = 32.87 xx 10^(-20) J`
(d) Energy of one photon (quantum) `= 32.27 xx 10^(-20) J`
Therefore, `32.27 xx 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 xx 10^(18)`