An experiment is conducted to investigate the photoelectriceffect. When light of

Question

An experiment is conducted to investigate the photoelectriceffect. When light of frequency 1.0 x 10 15 Hz isincident on a photocathode, electrons are emitted. Currentdue to these electrons can be cut off with a 1.0 V stoppingpotential. Light of frequency 1.5 x 10 15 Hzproduces a photoelectric current that can be cut off with a 3.0 Vstopping potential Calculate an experimental value of Planck's constant based onthis data Calculate the work function of the photocathode Will electrons be emitted from the photocathode when greenlight of wavelength 5.0 x 10 -7 meter is incidenton the photocathode? Justify answer. An experiment is conducted to investigate the photoelectriceffect. When light of frequency 1.0 x 10 15 Hz isincident on a photocathode, electrons are emitted. Currentdue to these electrons can be cut off with a 1.0 V stoppingpotential. Light of frequency 1.5 x 10 15 Hzproduces a photoelectric current that can be cut off with a 3.0 Vstopping potential Calculate an experimental value of Planck's constant based onthis data Calculate the work function of the photocathode Will electrons be emitted from the photocathode when greenlight of wavelength 5.0 x 10 -7 meter is incidenton the photocathode? Justify answer.

Explanation / Answer

Stopping potential V is related to max KE of electrons by eV = max KE = hf -W                         (W = work function) eV' = hf' - W e(V-V') = h(f-f') h = e(V-V')/(f-f') = 1.6*10^-19 (3-1)/(1.5*10^15 - 10^15) =6.4*10^-34 J.s W = hf - eV = 6.4*10^-34 *10^15 - 1.6*10^-19 * 1 = 4.8*10^-19 J = 3eV photons of green light has = 5*10^-7 m f = c/ = 3*10^8 /5*10^-7 = 6*10^14 Hz hf = 6.6*10^-34 * 6*10^14 /(1.6*10^-19) = 2.48 eV this is less than work function. so no electrons will beemitted.

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