|By Physicous (Physicous) on Monday, April 26, 2004 - 06:25 pm: Edit|
a sodium photoelectric surface with work funtion 2.3 eV is illuminated by electromagnetic radiation and emits electrons. the electrons travel toward a negatively charged cathode and complete the circuit as shown above. the potential difference supplied by the power supply is increased, and when it reaches 4.5 eV, no electrons reach the cathode.
a) For the electrons emitted from the sodium surface, calculate the following:
i. the maximum kinetic energy
ii. the speed at this maximum kinetic energy
b. calculate the wavelength of the radiation that is incident on the sodium surface.
c. calculate the minimum frequency of the light that will cause the photoemission from this sodium surface.
#5 here: http://kitsilano.vsb.bc.ca/students/ap/00-AP-Physics_FR.pdf
|By Tout (Tout) on Monday, April 26, 2004 - 11:49 pm: Edit|
The maximum potential energy equals the cutoff potential - 4.5eV.
Ek = 4.5eV = 1/2 mv^2. Isolate for v.
Energy of light = Ek + Work Function = 4.5eV + 2.3eV.
E = hf = hc/w = 6.8eV where w denotes wavelength.
Minimum frequency light must have energy equal to work function
E = hf = 2.3eV. Isolate for f.
|By Bfoleyx3 (Bfoleyx3) on Tuesday, April 27, 2004 - 05:20 pm: Edit|
thanks a lot.
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