JEE PYQ: Atomic Physics Question 52
Question 52 - 2019 (10 Apr Shift 2)
A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is:
[Given Planck’s constant $h = 6.6 \times 10^{-34}$ Js, speed of light $c = 3.0 \times 10^8$ m/s]
(1) $5 \times 10^{15}$
(2) $1.5 \times 10^{16}$
(3) $2 \times 10^{16}$
(4) $1 \times 10^{16}$
Show Answer
Answer: (1)
Solution
Energy of photon $E = \frac{hc}{\lambda}$. Number of photons emitted in 1 second: $n = \frac{P\lambda}{hc} = \frac{2 \times 10^{-3} \times 5 \times 10^{-7}}{2 \times 10^{-25}} = 5 \times 10^{15}$.
BETA
