A photoelectric tube is set up in a darkened corner of the lab: a clean metal cathode sealed in an evacuated glass envelope, a collector facing it, and a sensitive ammeter in the external circuit. The cathode's work function is listed on the manufacturer's data sheet as 3.10 eV. A mercury lamp fitted with a filter that passes its 245 nm ultraviolet line is aimed at the window, and the instant the lamp's shutter opens the ammeter needle lifts — electrons are leaving the metal and crossing the vacuum to the collector. The fastest electrons it releases are singled out in the usual way, by finding the retarding voltage that just silences the current. 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.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
φ = 3.1 eV — Work function of the cathode
λ = 245 nm — Wavelength of the ultraviolet light
Determine
(a)the energy of one incoming photon
(b)the maximum kinetic energy of an ejected electron
(c)the speed that corresponds to
(d)the de Broglie wavelength of an electron that fast
Step 1 of 4(a) · 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=λhc
Your values, in your units
E=(245nm)(6.62607e−34J⋅s)(299,792,000m/s)
Converted to base units
E=(2.45000e−07m)(6.62607e−34J⋅s)(299,792,000m/s)
Answer
E=5.0606eV
Carried onward at full precision, not this rounded figure.
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=Ephoton−ϕ
5.0606 eVcarried from step 1
Your values, in your units
KEmax=(8.10794e−19J)−(3.1eV)
Converted to base units
KEmax=(8.10794e−19J)−(4.96675e−19J)
Answer
KEmax=1.9606eV
Carried onward at full precision, not this rounded figure.
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=m2Ek
1.9606 eVcarried from step 2
Your values, in your units
v=(9.10938e−31kg)2(3.14119e−19J)
Answer
v=830.46km/s
Carried onward at full precision, not this rounded figure.
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=mvh
830.46 km/scarried from step 3
Your values, in your units
λ=(9.10938e−31kg)(830,458m/s)(6.62607e−34J⋅s)
Answer
λ=875.89pm
Carried onward at full precision, not this rounded figure.
Therefore each photon arrives with 8.11 × 10⁻¹⁹ J (5.06 eV), the fastest electrons keep 3.14 × 10⁻¹⁹ J (1.96 eV) once the work function is met, that is a speed of 8.30 × 10⁵ m/s, and such an electron travels with a de Broglie wavelength of 0.876 nm — some 280 times shorter than the light that freed it.
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 delivers one of them to 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.
