Question 172 marksVery Short AnswerSection B

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.

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 λ=200 nm\lambda = 200 \text{ nm}
  • Stopping potential V0=0.80 VV_0 = 0.80 \text{ V} (photocurrent zero at −0.80 V)

Photon energy:

  • E=hc/λ=1240 eV⋅nm÷200 nmE = hc/\lambda = 1240 \text{ eV} \cdot \text{nm} \div 200 \text{ nm} = 6.2 eV

Maximum kinetic energy:

  • Kmax⁡=eV0=0.80 eVK_{\max} = eV_0 = 0.80 \text{ eV}

Work function:

  • ϕ0=E−Kmax⁡=6.2−0.80\phi_0 = E - K_{\max} = 6.2 - 0.80 = 5.4 eV

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

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

Marking scheme

  1. 1.1 mark: photon energy E=hc/λ=1240/200=6.2 eVE = hc/\lambda = 1240/200 = 6.2 \text{ eV}.
  2. 2.½ mark: identifying Kmax⁡=eV0=0.80 eVK_{\max} = eV_0 = 0.80 \text{ eV}.
  3. 3.½ mark: ϕ0=6.2−0.80=5.4 eV\phi_0 = 6.2 - 0.80 = 5.4 \text{ eV} (accept 5.4 eV within rounding of hc).

Hint

ϕ0=hc/λ−eV0\phi_0 = hc/\lambda - eV_0; use hc=1240 eV⋅nmhc = 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.

Common mistakes

  1. 1.Using the −0.80 V sign directly and getting ϕ0=7.0 eV\phi_0 = 7.0 \text{ eV}; only the magnitude 0.80 eV is the stopping energy.
  2. 2.Forgetting to convert hc/λhc/\lambda correctly — E(eV)=1240/λ(nm)E(\text{eV}) = 1240/\lambda(\text{nm}) gives 6.2 eV for 200 nm.
  3. 3.Equating work function to the whole photon energy and ignoring Kmax⁡K_{\max}.

Analysis and explanation

Concept:

Einstein's photoelectric equation states hν=ϕ0+Kmax⁡h\nu = \phi_0 + K_{\max}, where the maximum kinetic energy of the ejected electrons is measured through the stopping potential, Kmax⁡=eV0K_{\max} = eV_0.


Method:

  • Convert the photon's energy using the handy relation E(eV)=1240/λ(nm)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.

Interesting facts

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

The convenient constant hc≈1240 eV⋅nmhc \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.

Question details

Q17 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Dual Nature of Radiation and Matter
Topic
Photoelectric effect — work function
Bloom's level
apply
Syllabus topic
Unit VII: Dual Nature of Radiation and Matter — Einstein's photoelectric equation; stopping potential and work function
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do you get photon energy from wavelength quickly?

Use E(eV)=1240/λ(nm)E(\text{eV}) = 1240 / \lambda(\text{nm}). For 200 nm this gives 1240/200=6.2 eV1240/200 = 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.