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A short H2 Physics revision video on H2 Physics 19 - Quantum Physics: Photoelectric Effect Calculation, built for quick recap before tutorial practice or exam revision.
Read through the explanation after watching, or jump straight to the step you want to replay.
Step 1 - State the problem
Ultraviolet light of wavelength two hundred and fifty nanometres strikes a metal surface with a work function of three point five electron-volts.
Step 1 - State the problem
We want to find the energy of each photon, check whether photoelectric emission occurs, and find the maximum kinetic energy of the emitted electrons.
Step 2 - Calculate the photon energy
The energy of a photon is E equals h f, which can also be written as h c over lambda.
Step 2 - Calculate the photon energy
Substituting: E equals six point six three times ten to the minus thirty-four times three point zero zero times ten to the eight, divided by two point five zero times ten to the minus seven.
Step 2 - Calculate the photon energy
The numerator is one point nine eight nine times ten to the minus twenty-five.
Step 3 - Convert to electron-volts and check threshold
Dividing gives E equals seven point nine six times ten to the minus nineteen joules.
Step 3 - Convert to electron-volts and check threshold
Converting to electron-volts: divide by one point six zero times ten to the minus nineteen to get approximately four point nine seven electron-volts.
Step 3 - Convert to electron-volts and check threshold
Since four point nine seven e V is greater than the work function of three point five e V, photoelectric emission does occur.
Step 4 - Apply Einstein's photoelectric equation
Einstein's photoelectric equation says the photon energy equals the work function plus the maximum kinetic energy of the emitted electron.
Step 4 - Apply Einstein's photoelectric equation
Rearranging: K E max equals E minus phi equals four point nine seven minus three point five, which gives one point four seven electron-volts.
Step 4 - Apply Einstein's photoelectric equation
Converting to joules: one point four seven times one point six zero times ten to the minus nineteen gives two point three five times ten to the minus nineteen joules.
Step 5 - Find the stopping potential
The stopping potential V s is the voltage needed to just stop the fastest photoelectrons.
Step 5 - Find the stopping potential
The work done by the electric field equals the maximum kinetic energy: e V s equals K E max.
Step 5 - Find the stopping potential
So V s equals K E max divided by e, which in electron-volts is simply one point four seven volts.
Step 5 - Find the stopping potential
Key point: increasing the intensity of the light increases the number of photoelectrons but does not change the maximum kinetic energy. Only increasing the frequency increases K E max.