In a photoelectric experiment, the emitter plate is irradiated with radiation of 200 nm. The photocurrent becomes zero when the collector plate potential is − 0.80 V. Calculate the work function (in eV) of the emitter.

Q: How do you get photon energy from wavelength quickly?

Use $E(\text{eV}) = 1240 / \lambda(\text{nm})$. For 200 nm this gives $1240/200 = 6.2 \text{ eV}$, avoiding a joule-to-eV conversion.

Dual Nature of Radiation and Matter

Photoelectric effect — work function

Official Answer

Lead: Use Einstein's photoelectric equation with the stopping potential to find the work function.

Given:

Wavelength $\lambda = 200 \text{ nm}$
Stopping potential $V_0 = 0.80 \text{ V}$ (photocurrent zero at −0.80 V)

Photon energy:

$E = hc/\lambda = 1240 \text{ eV} \cdot \text{nm} \div 200 \text{ nm}$ = 6.2 eV

Maximum kinetic energy:

$K_{\max} = eV_0 = 0.80 \text{ eV}$

Work function:

$\phi_0 = E - K_{\max} = 6.2 - 0.80$ = 5.4 eV

Result: The work function of the emitter is 5.4 eV.

photoelectric equationwork functionstopping potentialphoton energyhc = 1240 eV nm5.4 eVmaximum kinetic energyEinstein equation

Marking Scheme

11 mark: photon energy $E = hc/\lambda = 1240/200 = 6.2 \text{ eV}$ .
2½ mark: identifying $K_{\max} = eV_0 = 0.80 \text{ eV}$ .
3½ mark: $\phi_0 = 6.2 - 0.80 = 5.4 \text{ eV}$ (accept 5.4 eV within rounding of hc).

Hint

$\phi_0 = hc/\lambda - eV_0$ ; use $hc = 1240 \text{ eV} \cdot \text{nm}$ and stopping potential 0.80 V.

Quick Oral Answer

The 200 nm photon carries 6.2 eV; the 0.80 V stopping potential means the fastest electrons have 0.80 eV, so the work function is 6.2 minus 0.80, which equals 5.4 eV.

Analysis & Explanation

Concept:

Einstein's photoelectric equation states $h\nu = \phi_0 + K_{\max}$ , where the maximum kinetic energy of the ejected electrons is measured through the stopping potential, $K_{\max} = eV_0$ .

Method:

Convert the photon's energy using the handy relation $E(\text{eV}) = 1240 / \lambda(\text{nm})$ .
The stopping potential of 0.80 V means the most energetic electrons carry 0.80 eV.
Subtracting gives the minimum energy needed to free an electron — the work function.

Exam trap:

The collector potential is −0.80 V; its magnitude 0.80 V is the stopping potential. Do not carry the minus sign into eV₀.
Keep everything in eV to avoid converting to joules unnecessarily.

Real-world:

Work-function measurement like this is how sensor designers choose photocathode metals for photomultiplier tubes and night-vision devices — a low work function eases electron emission.

Common Mistakes

1Using the −0.80 V sign directly and getting $\phi_0 = 7.0 \text{ eV}$ ; only the magnitude 0.80 eV is the stopping energy.
2Forgetting to convert $hc/\lambda$ correctly — $E(\text{eV}) = 1240/\lambda(\text{nm})$ gives 6.2 eV for 200 nm.
3Equating work function to the whole photon energy and ignoring $K_{\max}$ .

Interesting Facts

Einstein received the 1921 Nobel Prize specifically for explaining the photoelectric effect, not for relativity.

The convenient constant $hc \approx 1240\ \text{eV}\cdot\text{nm}$ turns any wavelength in nm directly into photon energy in eV.

A 5.4 eV work function is high — typical metals range 2–5 eV, so this emitter needs deep-UV light (200 nm) to eject electrons.

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Frequently Asked Questions

How do you get photon energy from wavelength quickly?

Use $E(\text{eV}) = 1240 / \lambda(\text{nm})$ . For 200 nm this gives $1240/200 = 6.2 \text{ eV}$ , avoiding a joule-to-eV conversion.

Why is the stopping potential 0.80 V and not −0.80 V in the formula?

The negative sign only tells us the collector is retarding electrons. The maximum kinetic energy equals e times the magnitude of the stopping potential, i.e. 0.80 eV.

What does a 5.4 eV work function tell us?

It is the minimum energy to free an electron from the emitter; being fairly high, it requires ultraviolet light (like 200 nm) for photoemission to occur.