Light rays of wavelengths 6000 Å and of photon intensity 39.6 watt/m2 are incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be: (h = 6.64 x 10^-34 Js, velocity of light = 3 x 10^8 m/s)
Correct answer: C. 12 x 10^17
- A. 12 x 10^18
- B. 10 x 10^18
- C. 12 x 10^17
- D. 12 x 10^16
Explanation
Energy of one photon equals h c over lambda equals 6.64 × 10⁻³⁴ joule seconds times 3 × 10⁸ meters per second divided by 6000 angstrom equals 3.32 × 10⁻¹⁹ joule.Number of photons hitting one square meter each second equals intensity divided by photon energy equals 39.6 joule per second per square meter divided by 3.32 × 10⁻¹⁹ joule equals about 1.19 × 10²⁰.Only one percent of the photons cause emission so electrons per second per square meter equals 0.01 times 1.19 × 10²⁰ equals 1.19 × 10¹⁸.
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Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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