Consider a parallel beam of light of wavelength 600nm and intensity 100W (`m^-2)`. (a) Find the energy and linear momentum of each photon. (b) How many photons cross 1 `cm^2` area perpendicular to the beam in one second?

Explanation

(a) The energy of each photon `E = hc/ lambda `
` = ((4.14 xx (10^-15)eVs) xx (3 xx (10^8) m (s^-1))/600 xx (10^-9) m)= 2.07 e V. `
The linear momentum is
` p=E/c=(2.07eV / 3 xx (10^8)m (s^-1))= 0.69 xx (10^-8 eVs (m^-1)`.
(b) `The energy crossing 1 cm^2 in one second.`
`=(100 W(m^-2)) xx (1 (cm^2)) xx (1s) = 1.0 xx (10^-2)J.`
The number of photons making up this amount of
energy is
` n = (1.0 xx (10^-2)J/ (2.07)e V) = (1.0 xx (10^-2)/ ( 2.07 xx 1.6 xx (10^-19)) = 3.0 xx 10^16. `