The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength of the incident radiation is changed from 400 nm to 310 nm. The work function of the metal is:
Correct answer: C. 2.2 eV
- A. 0.9 eV
- B. 1.7 eV
- C. 2.2 eV
- D. 3.1 eV
Explanation
Let's first calculate the energy of the first radiation. Energy = hv/λ = (6.63x10-34 x 3x108)/ 400x10-9 = 4.9725x10-19 J = 4.9725x10-19 / 1.6 x10-19 = 3.1 ev The energy of the second radiation Energy = hv/λ = (6.63x10-34 x 3x108)/ 310x10-9 = 6.416x10-19 J = 6.416x10-19/1.6x10-19 = 4 ev K.E = Energy - Φ = 3.1 - Φ The second radiation will have twice K.E 2K.E = 4 - Φ K.E = (4 - Φ)/2 K.E = K.E (4 - Φ) / 2 = 3.1 - Φ 4 - Φ = 2(3.1 - Φ) 4 - Φ = 6.2 - 2Φ -Φ + 2Φ = 6.2 - 4 Φ = 2.2 ev
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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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