1) If light is shone on an anode, a current will be produced containing electron
ID: 1684031 • Letter: 1
Question
1) If light is shone on an anode, a current will be produced containing electrons with a particular maximum kinetic energy. Placing a potential across the metal such that it opposes the current will slow the electrons. You illuminate a cesium surface with light with wavelength 5230 Å and measure a stopping potential V0 of 0.23 V, and then change the wavelength to 4080 Å and measure a V0 of 0.90 V.a. What is the frequency of the 5230 Å wavelength light source?
b. What is the frequency of the 4080 Å wavelength light source?
c. For the electrons with 0.23V stopping potential, find the maximum kinetic energy the electrons will have when they are ejected from the anode.
d. For the electrons with 0.90V stopping potential, find the maximum kinetic energy the electrons will have when they are ejected from the anode.
e. Find an experimental value of Planck’s constant.
f. What is the work function of the anode?
Explanation / Answer
1) w1 = 5230 Å, V01 = 0.23 V, w2 = 4080 Å, V02 = 0.90 V. a. What is the frequency of the 5230 Å wavelength light source? f1 = c/w1 = 5.74*10^14 Hz b. What is the frequency of the 4080 Å wavelength light source? f2 = c/w2 = 7.35*10^14 Hz c. For the electrons with 0.23V stopping potential, find the maximum kinetic energy KE1 the electrons will have when they are ejected from the anode. KE1 = q*V01 = 0.23 eV d. For the electrons with 0.90V stopping potential, find the maximum kinetic energy KE2 the electrons will have when they are ejected from the anode. KE2 = q*V02 = 0.90 eV e. Find an experimental value of Planck’s constant h. hf1 - W = KE1 hf2 - W = KE2 h(f1 - f2) = KE1 - KE2 so h = (KE1 - KE2)/(f1 - f2) = 6.66*10^-34 Js f. What is the work function of the anode? W = hf1 - KE1 = 4.35*10^-19 J = 2.16 eV
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.