The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1   mW and wavelength of 1000   nm , in 0 . 1 second is x × 10 13 . The value of x is (Nearest integer) h = 6 . 63 × 10 - 34 Js ,   c = 3 . 00 × 10 8   ms - 1 :

Step-by-step solution

Energy emitted in 0 . 1 sec = 0 . 1   sec × 10 - 3   J   s - 1 = 10 - 4   J If 'n' photons of λ = 1000   nm are emitted, then 10 - 4 = nhc λ 10 - 4 = n × 6 . 63 × 10 - 34 × 3 × 10 8 1000 × 10 - 9 n = 5 . 02 × 10 14 = 50 . 2 × 10 13 n = 50 (nearest integer)