The work function of a metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of the emitted photo-electrons will be:
Correct answer: D. 1 x 10^6 m/ sec
- A. 10 m/sec
- B. 1 x 10^3 m/ sec
- C. 1 x 10^4 m/ sec
- D. 1 x 10^6 m/ sec
Explanation
Convert the wavelength to meters: 3000 Å = 3000 x 10-10 m = 3 x 10-7 m Calculate the energy of the incident photon (E) using the formula E = hc/λ, where h is Planck's constant (6.626 x 10-34 Js) and c is the speed of light (3 x 108 m/s). E = (6.626 x 10-34 Js * 3 x 108 m/s) / (3 x 10-7 m) = 6.626 x 10-19 J Convert the work function (Φ) from eV to Joules: 1 eV = 1.602 x 10-19 JCalculate the kinetic energy (KE) of the emitted electron using KE = E - Φ. KE = (6.626 x 10-19 J) - (1.602 x 10-19 J) = 5.024 x 10-19 JCalculate the velocity (v) of the electron using KE = 1/2 mv2, where m is the mass of an electron (9.109 x 10-31 kg). v = √(2KE/m) = √((2 * 5.024 x 10-19 J) / (9.109 x 10-31 kg)) ≈ 1.05 x 106m/s
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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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