(a) Here, power of transmitter, P = 10 kW = 104 W
∴ Total energy emitted per second = P × t = 104 × 1 = 104 J
Energy of each photon,
E = hυ =

hc

λ

=

6.63×10−34×3×108

500

If n is the number of photons emitted, then nE = 104
or n =

104

E

=

104×500

6.63×3×10−26

= 2.51 × 1031
We see that the energy of a radio photon is exceedingly small and the number of photons emitted per second in a radio beam is enormously large. Therefore, negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave as continuous.
(b) Here, area of the pupil, A = 0.4 cm2 = 0.4 × 10–4m2, u = 6 × 1014 Hz
Intensity = 10–10Wm–2
Energy of a photon is given by, E = hυ = 6.63 × 10–34 × 6 × 1014 J ≈ 4 × 10–19 J.
If n = number of photons falling per sec per unit area, the energy per unit area per sec due to these photons = total energy of n photons = n × 4 × 10–19Jm–2
Since, intensity = energy per unit area per second
∴ 10–10 = n × 4 × 10–19
or n =

10−10

4×10−19

= 2.5 × 108m−2s−1
∴ Number of photons entering the pupil per second = n × area of the pupil = 2.5 × 108 × 0.4 × 10–4s–1 = 104s–1.
Though this number is not large as in part (a) above, it is large enough for us to ‘sense’ or ‘count’ the individual photons by our eye.