JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Atomic Structure

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 \times 1 0^{13}

. The value of

x

is ____. (Nearest integer)
(

h = 6.63 \times 1 0^{- 34} Js, c = 3.00 \times 1 0^{8} ms^{- 1}

)

Answer: 50

Solution

The total energy delivered in

0.1 s

E = P t = 1 0^{- 3} \times 0.1 = 1 0^{- 4} J

. Each photon of wavelength

1000 nm

carries

ε = \frac{h c}{λ} = \frac{6.63 \times 1 0 ^{- 34} \times 3.00 \times 1 0 ^{8}}{1000 \times 1 0 ^{- 9}} = 1.989 \times 1 0^{- 19} J .

The number emitted is

N = \frac{E}{ε} = \frac{1 0 ^{- 4}}{1.989 \times 1 0 ^{- 19}} = 5.03 \times 1 0^{14} = 50.3 \times 1 0^{13},

x = 50

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Integer answer