Photoelectric effect through to the electron's own wavelength

SPH4U Grade 12 Physics · Revolutions in Modern Physics

A clean metal cathode with a work function of 3.10 eV is illuminated with 245 nm ultraviolet light from a mercury lamp, and the fastest electrons it releases are collected in an evacuated tube. Find the energy of one incoming photon, the maximum kinetic energy of an ejected electron, the speed that corresponds to, and the de Broglie wavelength of an electron moving that fast.

Step 1 of 4 · solve for Photon energy

Light arrives in lumps, and E = hc/λ is the size of one lump. Brightness changes how many arrive per second, never how big each one is — which is the whole reason the classical picture failed.

Rearranged for E
E=hcλE = \tfrac{h c}{\lambda}
Your values, in your units
E=hc(245 nm)E = \tfrac{h c}{\left( 245\ \text{nm} \right)}
Converted to base units
E=hc(2.45000e07 m)E = \tfrac{h c}{\left( 2.45000e-07\ \text{m} \right)}
Answer
E=5.06E = 5.06

Carried onward at full precision, not this rounded figure.

Open the Photon Energy from Wavelength (E = hc/λ) solver →

Step 2 of 4 · solve for Max kinetic energy

One photon, one electron, no instalments. Pay the work function first to get the electron out of the metal; whatever is left over it keeps as kinetic energy.

Rearranged for KE
KEmax=EphotonϕKE_{max} = E_{photon} - \phi
5.06 eVcarried from step 1
Your values, in your units
KEmax=(8.10794e19 J)(3.1 eV)KE_{max} = \left( 8.10794e-19\ \text{J} \right) - \left( 3.1\ \text{eV} \right)
Converted to base units
KEmax=(8.10794e19 J)(4.96675e19 J)KE_{max} = \left( 8.10794e-19\ \text{J} \right) - \left( 4.96675e-19\ \text{J} \right)
Answer
KEmax=1.96KE_{max} = 1.96

Carried onward at full precision, not this rounded figure.

Open the Photoelectric Effect solver →

Step 3 of 4 · solve for Speed

Turn that energy into a speed with the ordinary ½mv². At 0.28% of light speed relativity is not yet needed, though a hundred times this energy would change the answer.

Rearranged for v
v=2Ekmv = \sqrt{\tfrac{2 E_k}{m}}
1.96 eVcarried from step 2
Your values, in your units
v=2(3.14119e19 J)(9.10938e31 kg)v = \sqrt{\tfrac{2 \, \left( 3.14119e-19\ \text{J} \right)}{\left( 9.10938e-31\ \text{kg} \right)}}
Answer
v=830,000v = 830{,}000

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 4 of 4 · solve for De Broglie wavelength

The last step turns the argument around. Having treated a wave as a particle for three steps, treat the particle as a wave and ask what its wavelength is.

Rearranged for λ
λ=hmv\lambda = \tfrac{h}{m v}
830,000 m/scarried from step 3
Your values, in your units
λ=h(9.10938e31 kg)(830,458 m/s)\lambda = \tfrac{h}{\left( 9.10938e-31\ \text{kg} \right) \, \left( 830{,}458\ \text{m/s} \right)}
Answer
λ=876\lambda = 876

Carried onward at full precision, not this rounded figure.

Open the De Broglie Wavelength solver →

Why this order

The order tells the story of 1905 and 1924 in four lines. Step 1 chops a continuous beam into photons; step 2 spends one of them on one electron; step 3 reads the leftovers as ordinary motion; step 4 hands the wave nature back to the electron. The step that must come second is the subtraction, and the reason is the thing Einstein saw: an electron cannot save up. A faint beam of 5.06 eV photons frees electrons immediately, while an arbitrarily bright beam of 2.00 eV photons frees none at all, ever, no matter how long you leave it running. That threshold behaviour is what no wave theory could produce, and it is why the work function is subtracted from a single photon's energy rather than from the beam's.

Two traps, both bookkeeping. The first is the electron volt: 3.10 eV has to become 4.967 × 10⁻¹⁹ J before it can be subtracted from anything, and mixing the two units is the single most common way to lose the question. The second is confusing the photon's 245 nm with the electron's 0.876 nm at the end. They are wavelengths of entirely different things, and the electron's is roughly 280 times shorter even though the electron carries barely a third of the photon's energy — because λ = h/mv depends on momentum, and a slow massive particle has far more momentum per joule than a photon does. That mismatch is not a curiosity; it is the reason an electron microscope resolves atoms and an optical one never will.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.