A point of of 6 watts emits monochromatic light of wavelength 5000dot A- The number of photons striking normally per second per unit area of the surface distant 5 m from the will be4-82 space times space 10-16-4-824-82 space times space 10-6-4-82 space times space 10-4-

Question

Toppr Admin · Accepted Answer

The photons will be emitted symmetrically in all directions.
Power of Source = $6 W$

$E n e r g y = n h ν = \frac{n h c}{λ}$ where $n$ is no.of $p h o t o n s$ emitted per second.

Soving the above equation we get n.

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

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

So, the answer is option (D).

A point of source of 6 watts emits monochromatic light of wavelength 5000˙A. The number of photons striking normally per second per unit area of the surface distant 5 m from the source will be4.82 × 10164.824.82 × 10−64.82 × 10−4

A point of source of 6 watts emits monochromatic light of wavelength $5000 ˙ A$ . The number of photons striking normally per second per unit area of the surface distant 5 m from the source will be
$4.82 \times 10^{16}$
$4.82$
$4.82 \times 10^{- 6}$
$4.82 \times 10^{- 4}$