Wavelength of radiation emitted = 616 nm = 616 × 10–9 m (Given)

(a) Frequency of emission (v)
c = c/$\mathrm{\lambda }$
Where, c = velocity of
radiation λ = wavelength ofradiation

Substituting the values in the given expression of (v):
$\mathrm{v} = \frac{3.0 \mathrm{x} {10}^8 \mathrm{m}/\mathrm{s}}{616 \mathrm{x} {10}^{-9} \mathrm{m}}$
= 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 m/s) (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
$= \frac{2\mathrm{J}}{32.27 \mathrm{x} {10}^{-20} \mathrm{J}}$
$= 6.19 \times {10}^{18}$
$= 6.2 \times {10}^{18}$