Exams › JEE Main › Physics

A sodium street lamp rated at 200 W produces yellow radiation of wavelength 0.6 μm. If only 25% of the electrical power is converted into light, how many yellow photons are emitted each second?

1. 1.5 × 10²⁰
2. 6 × 10¹⁸
3. 6 × 10²⁰
4. 3 × 10¹⁹

Correct answer: 1.5 × 10²⁰

Solution

Useful light power = 25% of 200 W = 50 W. Energy per photon = hc/lambda = (6.63e-34*3e8)/(0.6e-6) = 3.3e-19 J. Photons per second = 50/3.3e-19 = 1.5e20.

Related JEE Main Physics questions

• A star is sending out light of wavelength 5000 Å and is moving toward the Earth with a speed of 1.50 × 10⁶ m/s. What is the shift in the wavelength of the light observed on Earth?
• A beam of radiation carrying energy E is incident perpendicularly on a perfectly reflecting surface. What momentum is imparted to the surface? (C = speed of light)
• Two identical photocathodes are illuminated by light of frequencies f1 and f2. If the emitted photoelectrons have speeds v1 and v2 respectively, then
• An X-ray tube is run with a potential difference of 15 kV. What is the maximum possible speed of the electrons that hit the target?
• In a photoelectric setup, monochromatic light of wavelength λ ejects electrons, and the maximum speed of the emitted electrons is v. If the incident wavelength is changed to 5λ/4, the speed of the emitted electrons will be
• An electron and a photon have the same energy E. If their wavelengths are denoted by λₑ and λₚ respectively, which relation between them is correct?

⚔️ Practice JEE Main Physics free + battle 1v1 →