Question

A point of source of 6 watts emits monochromatic light of wavelength $5000\dot{A}$. The number of photons striking normally per second per unit area of the surface distant 5 m from the source will be

• A. $4.82$
• B. $4.82\times {10}^{-4}$
• C. $4.82\times {10}^{-6}$
• D. $4.82\times {10}^{16}$

Answer 1

Answer: (D)

$4.82\times {10}^{16}$

The photons will be emitted symmetrically in all directions.

Power of Source = $6W$

$Energy=nh\nu =\frac{nhc}{\lambda }$ where $n$ is no.of $photons$ emitted per second.

Soving the above equation we get n.

The no.of photons falling on $1m^2$ area is equal to $\frac{n}{4\pi d^2}$ where $d$ is the distance of the point.

$\Rightarrow$No.of photons=$\frac{1.51\times {10}^{19}}{4\pi (5{)}^2}=4.82\times {10}^{16}$

So, the answer is option (D).