# File - King's Senior Science

```•Set up the circuit to supply an opposing
Voltage to measure the (maximum) kinetic
energy of the electrons
•Adjust the variable voltage supply until the
current is zero. This voltage is called the cutoff voltage.
•Note how the cell is arranged – where it is
‘pushing’ electrons
•Measure the cut-off voltage for different
light frequencies and plot data on a graph.
•Convert Voltage to Kinetic energy
Ek
V= Energy / charge
so E = qV
Ek = eV
where e is the charge on 1 electron
1 electron volt =1 eV = 1.6 x 10-19 J
V
f
f
Kinetic energy vs. frequency:
Ek
Ek  hf  
-ϕ
fo
h
f
Terms &amp; Definitions:
Threshold frequency (fo) the minimum frequency of light
that results in the emission of photoelectrons
Work function (ϕ) the minimum amount of work needed to
eject an electron from the surface ϕ = h fo
Stopping potential (V) equivalent to the maximum kinetic
energy of the ejected photoelectrons (Ek)
A beam of 640nm red light shines onto a polished
copper disc.
1. Calculate the energy of each photon
E=hf=hc/=3.11x10-19J
2.Write this energy in eV
1eV=1.6x10-19J  1.9eV
The work function of copper is 1.28x10-18J
3.Figure out whether an electron will be emitted
Φ = 1.28 x 10-18J. Ei = 3.11x10-19J is about &frac14;
of what's needed NO electron emitted
1.24x10-18J. Still not quiet enough
5.What is 160 nm Light? UV
Blue light wavelength 429nm is shone on a metal
surface with threshold frequency 3.0x1014Hz.
What is the maximum velocity of the electrons
(mass=9.1x10-31kg)?
Step one find the frequency of the light.
Step two calc the photon energy
4.6x10-19J
Use hfo=Φ, Ek=hf- Φ, Ek=&frac12;mv2 to get v
v=7.6x105ms-1
7.0x1014Hz
A sodium surface is illuminated with light of wavelength 300nm. The work
function for sodium metal is 2.46 eV. Find:
a) The Maximum kinetic energy of the ejected photoelectrons
b) The threshold frequency (&amp; wavelength) for sodium
Solution a)
The energy of the photons in the illuminating light beam is E = hf = hc/ 
E = (6.626 x 10-34 J.s)(3.00 x 108 m/s) / 300 x 10-9 m
= 6.63 x 10-19 = 6.63 x 10-19 J / 1.60 x 10-19 J/eV = 4.14 eV
Using equation Ek max = hf - Φ
we have Ek max = 4.14 -2.46 = 1.68 eV = 2.69 x 10-19 J
Solution b)
At the Threshold frequency Ek max = 0. So Φ = hf o
since Φ = 2.46 eV = (2.46eV) (1.60 x 10-19 J/eV) = 3.94 x 10-19 J
Then fo = Φ /h
= 3.94 x 10-19 J / 6.626 x 10-34 J.s
= 5.95 x 10 14Hz
= 505 nm
This is in the green region of the visible spectrum
What did you take in?
Try and answer these without looking back at