A sample of potassium metal (in a vacuum) is illuminated by a light source as sh
ID: 2301894 • Letter: A
Question
A sample of potassium metal (in a vacuum) is illuminated by a light source as shown above. The photoelectrons (i.e. electrons emitted from the potassium when exposed to light) are collected on plate P, and the electron current is measured by the meter G. The work function of potassium is 2.0 eV.
a) Initially red light (700 nm) is shown on the metal, but the wavelength of the light is slowly decreased. At what wavelength should a person first read a current on the meter?
b) As the wavelength is decreased until the light is violet (400 nm) will the current continue to increase? Assume that the intensity of the light stays constant as you decrease its wavelength.
c) Find the kinetic energy and the speed of the electrons emitted when the light source is at 400 nm.
Explanation / Answer
a) E = Wo + KE
initially KE = 0
E = Wo
h*c/lamda = Wo
lamda = hc/Wo
= 6.626*10^-34*3*10^8/(2*1.6*10^-19)
= 621.18 nm
b) no. magnitude of current dowen not depend on wavelength. it depends on intensity.
c) E = Wo + KE
KE = E - Wo
= h*c/lamda - wo
= 6.626*10^-34*3*10^8/(400*10^-9) - 2 eV
= 4.97*10^-19 J - 2 eV
= 3.1 eV - 2 eV
= 1.1 eV <<<<<<<-Answer
KE = 0.5*m*v^2
v = sqrt(2*KE/m)
= srt(2*1.1*1.6*10^-19/9.1*10^-31)
= 6.219*10^5 m/s <<<<<<-Answer
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.