Question

A 150 watt bulb emits monochromatic light of wavelength 662 nm. How many number of photons emitted per second by the bulb? (Planck's constant, \(\mathrm{h} = 6.62 \times 10^{- 34} \mathrm{~J} \mathrm{~s}, \mathrm{~c} = 3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{- 1}\))

• A. \(4 \times 10^{20} \mathrm{~s}^{-1}\)
• B. \(3 \times 10^{20} \mathrm{~s}^{-1}\)
• C. \(6 \times 10^{20} \mathrm{~s}^{-1}\)
• D. \(5 \times 10^{20} \mathrm{~s}^{-1}\)
• E. \(1 \times 10^{20} \mathrm{~s}^{-1}\)

Hint:

Use \(\lambda\) in meters. \(1 \text{ nm} = 10^{-9} \text{ m}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Power (P) = Energy emitted per second. Energy of one photon \(E = \frac{hc}{\lambda}\). Number of photons \(n = \frac{P}{E}\).

Step 2: Detailed Explanation:
\(\lambda = 662 \text{ nm} = 662 \times 10^{-9} \text{ m}\).
\(E = \frac{(6.62 \times 10^{-34})(3 \times 10^{8})}{662 \times 10^{-9}} = \frac{1.986 \times 10^{-25}}{6.62 \times 10^{-7}} = 3 \times 10^{-19} \text{ J}\).
\(n = \frac{150}{3 \times 10^{-19}} = 5 \times 10^{20} \text{ s}^{-1}\).

Step 3: Final Answer:
The number of photons emitted per second is \(5 \times 10^{20} \text{ s}^{-1}\).