Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons. |
Part A What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect? Express your answer using three significant figures.
SubmitMy AnswersGive Up Part B What is the wavelength of this light? Express your answer using three significant figures.
SubmitMy AnswersGive Up Part C This question will be shown after you complete previous question(s). Part D If titanium is irradiated with light of 233 nm, what is the maximum possible kinetic energy of the emitted electrons? Express your answer using three significant figures.
SubmitMy AnswersGive Up |
Part A :
minimum energy to emit electrons E = 6.94×10−19 J
E = hv
h = 6.625*10^-34 j.s
v= frequency = ?
(6.94*10^(−19)) = (6.625*10^(-34))*v
v = 1.05*10^15 s-1
Part B: wavelength of this light,
C = vL
c= light velocity = 3*10^8 m/s
v = frequency = 1.05*10^15 s-1
L = wavelength = ?
3*10^8 = (1.05*10^15)*L
L = 2.86*10^(-7) m = 286 nm
Part D
from photoelectric effect
hv = hv0 + K.E
K.E = (6.625*10^(-34)*(3*10^8/(233*10^(-9)))-(6.625*10^(-34)*1.05*10^15)
K.E = 1.57*10^-19 joule.
Get Answers For Free
Most questions answered within 1 hours.