The work function of a certain metal is 1.85 eV.
a. What is the value of the work function in units of Joules?
= 2.964 e -19 J
b. What is the value of f0, the lowest frequency of light that can remove electrons from this metal?
Formula: KE_max = hf - w = hf - hf0
But I only know w and h (a constant). I can't solve for anything else =/
I'm missing a piece of the puzzle. Can someone help me?
c. If light with a wavelength of 350 nm is incident on this material, what is the greatest kinetic energy of an electron ejected from the metal?
d. What voltage must be applied in order to stop the photoelectric current produced by the 350 nm light?
I am so confused ._.
Thank you!!!!
a. What is the value of the work function in units of Joules?
= 2.964 e -19 J
b. What is the value of f0, the lowest frequency of light that can remove electrons from this metal?
Formula: KE_max = hf - w = hf - hf0
But I only know w and h (a constant). I can't solve for anything else =/
I'm missing a piece of the puzzle. Can someone help me?
c. If light with a wavelength of 350 nm is incident on this material, what is the greatest kinetic energy of an electron ejected from the metal?
d. What voltage must be applied in order to stop the photoelectric current produced by the 350 nm light?
I am so confused ._.
Thank you!!!!
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a) W = 1.85 x 1.60E-19 = 2.96E-19J (not 2.964 e -19 J)
b) When the lowest frequency (f0) of light is used, the electron's kinetic energy is zero. The photon has only just enough energy to remove the electron (the work function), with none left over to give the electron any KE. So W= hf0
f0 = W/h = 2.96E-19/(6.626E-34) = 4.47E14Hz
c) Find photon energy. You could find f using f=c/lambda, then use E=hf. Or just use E = hc/lambda.
E= hc/lambda =(6.626E-34)x(3.00E8) / (350E-9) = 5.68E-19J
KEmax = E -W = 5.68E-19 - 2.96E-19 = 2.72E-19J
d) If we express the electrons max KE in units of eV, the value corresponds to the stopping potential.
KEmax = 2.72E-19J = 2.72E-19 / 1.6-E-19 eV = 1.7eV
This means the voltage which will stop the current (the stopping potential) is 1.7V
b) When the lowest frequency (f0) of light is used, the electron's kinetic energy is zero. The photon has only just enough energy to remove the electron (the work function), with none left over to give the electron any KE. So W= hf0
f0 = W/h = 2.96E-19/(6.626E-34) = 4.47E14Hz
c) Find photon energy. You could find f using f=c/lambda, then use E=hf. Or just use E = hc/lambda.
E= hc/lambda =(6.626E-34)x(3.00E8) / (350E-9) = 5.68E-19J
KEmax = E -W = 5.68E-19 - 2.96E-19 = 2.72E-19J
d) If we express the electrons max KE in units of eV, the value corresponds to the stopping potential.
KEmax = 2.72E-19J = 2.72E-19 / 1.6-E-19 eV = 1.7eV
This means the voltage which will stop the current (the stopping potential) is 1.7V
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