Solveeit Logo
Login

Question

Question: A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency then the number of p...

A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency then the number of photons emitted by the bulb in 20 seconds are (1eV = 1.6 × 10–19 J, hc = 12400 eV Å)

A

2 × 1018

B

1018

C

1021

D

2 × 1021

Answer

2 × 1021

Explanation

Solution

Power =40×80100=n×6.62×10−34×3×108620×10−9×20⇒n=2×1021= 40 \times \frac{80}{100} = \frac{n \times 6.62 \times 10^{- 34} \times 3 \times 10^{8}}{620 \times 10^{- 9} \times 20} \Rightarrow n = 2 \times 10^{21}