Nicolas A. answered • 09/14/19
Stanford Grad for Math, Science, and SAT/ACT
The energy E in one photon of light with frequency f is E = hf, where h = 6.63×10-34 J•s is Planck's constant.
First calculate the minimum energy required to eject the electron from the metal. The threshold frequency is minimum frequency of light with enough energy to eject the electron.
E = hf
E = (6.63×10-34 J•s) * (2.11×1014 s−1)
E = 1.40×10-19 J
Note: E = hf is often written E = hν (ν is the Greek symbol nu). The f and ν both mean frequency, but I used f to avoid confusion with v for velocity later on.
Now calculate the energy of the light that strikes the metal. The wavelength λ = 4.97×10−7 m is given instead of the frequency, so first calculate the frequency using λf = c, or f = c/λ, where c = 3.00×108 m/s is the speed of light. Then calculate the energy.
f = c/λ
f = (3.00×108 m/s) / 4.97×10−7 m
f = 6.04×1014 s−1
Then plug into E = hf
E = hf
E = (6.63×10-34 J•s) * (6.04×1014 s−1)
E = 4.00×10-19 J
The energy of the light (4.00×10-19 J) is more than the minimum needed (1.40×10-19 J) so it will eject the electron. Subtract the energies to see that the 4.00×10-19 J − 1.40×10-19 J = 2.60×10-19 J remaining energy is converted into kinetic energy of the ejected electron. Calculate its velocity using E = ½mv2, or v = √(2E/m), with electron mass m = 9.11×10-31 kg.
v = √(2*E/m)
v = √(2*(2.60×10-19 J / 9.11×10-31 kg))
v = 7.56×105 m/s
The ejected electron's velocity is 7.56×105 m/s.
Note: It's possible to use a single conservation-of-energy equation to solve for the electron's velocity, but some intuition for what's happening may be lost.
Elight = Ethreshold + Ekinetic
hflight = hfthreshold + ½mv2
hc/λlight = hfthreshold + ½mv2
hc/λlight − hfthreshold = ½mv2
√(2h(c/λlight − fthreshold)/m) = v
