An isotropic point source emits light with wavelength 500 nm. The radiation power of the source is P = 10 W . Find the number of photons passing through unit area per second at a distance of 3 m from the source.

5.92 x 10¹⁷ / m² s

2.23 x 10¹⁷ / m² s

2.23 x 10¹⁸ /m² s

5.92 x 10¹⁸ /m² s

Solution

2.23 x 10¹⁷ / m² s

Given:-

λ = 500 nm = 500 x 10^-9 m

Radiation power P = 10 W

As P = n_o $\frac{h c}{λ}$

n_o = $\frac{P λ}{hc}$ ......(i)

where, n_o is number of photons per second.

At distance r from point source, number of photons/area/time

n' = $\frac{n_{o}}{4 π r^{2}}$

n' = $\frac{P λ}{hc . 4 {πr}^{2}}$ [ from eq. (i) ]

= $\frac{10 \times 500 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8} \times 4 π {(3)}^{2}}$

= 2.23 × 10¹⁷ m/s

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