The work function for potassium is 2.25eV. A beam with a wavelength of 400nm and
ID: 2257126 • Letter: T
Question
The work function for potassium is 2.25eV. A beam with a wavelength of 400nm and an intensity of 10^-9W/m^2. Find
a) the maximum kinetic energy of the photelectrons.
b) the number of electrons emitted per meter squared per second from the surface, assuming 3% of the incident photons are effective in ejecting electrons.
for b) if the work function is the minimum amount of energy required to shoot out one electron,
why can't the problem be solved by going
10^-9 * (0.03) / work function = number of electron omitted?
Prefer answer to have work shown with brief explanations please
Explanation / Answer
a) Energy of incident photon = hc/lambda = 6.64*10^-34*3*10^8/ 400*10^-9 = 4.97*10^-19 J
Energy in joule / e = Energy in eV = 3.1 eV
The extra energy of photon is given to electron as their KE. This may be less depending on the structure. The max possible is (E - WF)
Energy of ejected electron = 3.1 - 2.25 = 0.85eV = 1.368*10^-19 J
b) Power = 10^-9 J/sec
Number of photom = power / energy of 1 photon = 2.013*10^9
Effective photon = 0.03*2.103*10^-9 J = 6.038*10^7
Number of electroms ejected = 6.038*10^7
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.