K_max = h ν - W, where K_max is maximum KE of emitted electron, hν is photon energy, W is work function of metal. Threshold ν₀ = W/h. Stopping potential V₀ satisfies eV₀ = K_max.
-- NCERT Class 12 Physics, Ch. 11, p. 279Einstein Photoelectric Equation
8 MCQs1 revision card9-step worked example
Source: NCERT Dual Nature of Matter and RadiationOfficial key: NTA-verifiedLast updated: 26 Sep 2026
Einstein Photoelectric Equation, explained for NEET
The equation is one line long and the marks are lost in one term. A photoelectric stem gives you a wavelength and a work function; the hand writes K_max = hν, the answer comes out clean and large, and it matches a distractor that was placed there for exactly that reason. The work function is not a correction — it is the energy the metal charges to release the electron at all, and it is subtracted every single time.
Einstein's equation, stated in NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter) at printed page 281: K_max = hν − W. A photon of frequency ν is absorbed whole by one electron. Part of that energy, W, pays the escape cost; whatever survives is kinetic energy, and the most any electron can keep is K_max. Electrons deeper in the metal pay more than W and emerge slower — which is why the equation gives a maximum, not the energy of every emitted electron.
Two consequences follow, and both are examinable. First, emission requires hν > W, so the threshold frequency is ν₀ = W/h. Below it, nothing is emitted no matter how intense the beam — intensity delivers more photons, not more energetic ones. Second, the stopping potential V₀ measures K_max directly: eV₀ = K_max = hν − W. That bridge is where most stems live. You are rarely asked for K_max in joules; you are asked for V₀ in volts, or for W in eV, and the conversion between the two is the eV₀ step.
For NEET the arithmetic is light but the setup is not. Work function is usually quoted in eV while photon energy arrives as hc/λ in joules — mixing the two units mid-line produces answers off by 1.6 × 10⁻¹⁹ that still look plausible. Convert first, subtract second.
Watch out: if a question gives you a work function or a threshold, it is telling you W is non-zero and expects to see it subtracted. A stem that supplies W and an answer that never used it is a wrong answer, however tidy the number.
Can you answer these Einstein Photoelectric Equation MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In Einstein's photoelectric equation, the quantity W represents:
Show answer and why every option is right or wrong
Answer: C. W is the work function — the minimum energy needed to free an electron from the metal, as defined with the equation in NCERT Class 12 Physics Chapter 11, printed page 281.
Why A is wrong: A is wrong because that quantity is K_max, which is what remains after W has been paid, not W itself.
Why B is wrong: B is wrong because the photon energy is hν; W is a property of the metal and is independent of the incident light.
Why D is wrong: D is wrong because eV₀ equals K_max, not W; confusing these is the same slip that produces K_max = hν.
MCQ 2Direct ApplicationPractice
Light of energy 5.0 eV falls on a metal of work function 2.0 eV. The maximum kinetic energy of the emitted photoelectrons is:
Show answer and why every option is right or wrong
Answer: A. K_max = hν − W = 5.0 − 2.0 = 3.0 eV, the direct subtraction stated in NCERT Class 12 Physics Chapter 11, printed page 281.
Why B is wrong: B is wrong because it reports hν alone — the work function has been dropped, which is precisely the error the given value of W is there to catch.
Why C is wrong: C is wrong because W has been added instead of subtracted; the metal charges energy, it does not supply it.
Why D is wrong: D is wrong because it divides the photon energy by 2 rather than subtracting W; no step in the equation involves halving.
MCQ 3Easy RecallPractice
The threshold frequency ν₀ of a metal is related to its work function W by:
Show answer and why every option is right or wrong
Answer: B. At threshold K_max = 0, so hν₀ = W, giving ν₀ = W/h — stated with Einstein's equation in NCERT Class 12 Physics Chapter 11, printed page 282.
Why A is wrong: A is wrong because multiplying W by h gives units of J²·s, not hertz; the relation is a quotient.
Why C is wrong: C is wrong because it inverts the ratio: h/W has units of (J·s)/J = s, a time rather than a frequency, so a dimensional check rejects it immediately.
Why D is wrong: D is wrong because dividing by c as well converts the result to a wave number (1/m), not a frequency; that form gives 1/λ₀, not ν₀.
MCQ 4Direct ApplicationPractice
Photons of energy 4.5 eV eject electrons from a metal whose work function is 1.9 eV. The stopping potential required to reduce the photocurrent to zero is:
Show answer and why every option is right or wrong
Answer: D. eV₀ = K_max = hν − W = 4.5 − 1.9 = 2.6 eV, so V₀ = 2.6 V — the stopping-potential bridge given in NCERT Class 12 Physics Chapter 11, printed page 282.
Why A is wrong: A is wrong because it equates eV₀ with the full photon energy, dropping the work function entirely.
Why B is wrong: B is wrong because it reports the work function as though V₀ measured the escape cost; V₀ measures what is left over after that cost.
Why C is wrong: C is wrong because the two energies have been added rather than subtracted.
MCQ 5Concept TrapPractice
Monochromatic light of frequency ν, with hν less than the work function W of a metal, is shone on that metal. The intensity of the light is then doubled. The result is:
Show answer and why every option is right or wrong
Answer: C. Emission requires hν > W; below the threshold frequency no emission occurs at any intensity, since each electron absorbs one photon whose energy is fixed by ν — NCERT Class 12 Physics Chapter 11, printed page 282.
Why A is wrong: A is wrong because intensity sets the number of photons arriving per second, not the energy of each one; K_max depends only on ν.
Why B is wrong: B is wrong because it assumes one electron absorbs two photons — the equation is built on single-photon absorption, so the energy term is hν and never 2hν.
Why D is wrong: D is wrong because it describes what happens above threshold; here hν < W, so no electron receives enough energy to escape and the rate is zero either way.
MCQ 6CalculationPractice
When a metal is illuminated by light of frequency ν, the stopping potential is V₀. When the frequency is raised to 2ν, the stopping potential becomes:
Show answer and why every option is right or wrong
Answer: C. eV₀ = hν − W and eV₀′ = 2hν − W = (hν − W) + hν, so V₀′ = V₀ + hν/e — applying the equation at both frequencies, NCERT Class 12 Physics Chapter 11, printed page 282.
Why A is wrong: A is wrong because doubling ν doubles only hν, not hν − W; the constant W means V₀ is not proportional to ν.
Why B is wrong: B is wrong because V₀′ − 2V₀ = W/e, which is positive, so V₀′ exceeds 2V₀ by W/e rather than falling short of it.
Why D is wrong: D is wrong because raising the frequency increases the leftover kinetic energy; halving the stopping potential inverts the dependence.
MCQ 7CalculationPractice
A metal has a threshold frequency of 6.0 × 10¹⁴ Hz. It is illuminated by light of frequency 9.0 × 10¹⁴ Hz. Taking h = 6.6 × 10⁻³⁴ J·s, the maximum kinetic energy of the emitted photoelectrons is:
Show answer and why every option is right or wrong
Answer: B. W = hν₀, so K_max = h(ν − ν₀) = 6.6 × 10⁻³⁴ × 3.0 × 10¹⁴ = 1.98 × 10⁻¹⁹ J — NCERT Class 12 Physics Chapter 11, printed page 282.
Why A is wrong: A is wrong because it evaluates hν alone (6.6 × 10⁻³⁴ × 9.0 × 10¹⁴), using the threshold only to confirm emission and then discarding W.
Why C is wrong: C is wrong because it evaluates hν₀, the work function itself, rather than the surplus above it.
Why D is wrong: D is wrong because it uses h(ν + ν₀); the threshold term is subtracted, since it represents an energy cost.
MCQ 8Concept TrapPractice
A graph of stopping potential V₀ against incident frequency ν is plotted for a given metal. The slope and the magnitude of the intercept on the V₀-axis correspond respectively to:
Show answer and why every option is right or wrong
Answer: A. Rearranging eV₀ = hν − W gives V₀ = (h/e)ν − W/e, a straight line of slope h/e with V₀-intercept −W/e — NCERT Class 12 Physics Chapter 11, printed page 282.
Why B is wrong: B is wrong because it omits the division by e; the equation must be divided throughout by the electronic charge before V₀ stands alone.
Why C is wrong: C is wrong because it inverts the slope, which would be the gradient of a ν-versus-V₀ plot, and multiplies the intercept by e instead of dividing.
Why D is wrong: D is wrong because the slope is right but the intercept is not; dividing eV₀ = hν − W by e gives W/e, so multiplying by e reverses the required operation.
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Einstein Photoelectric Equation: quick recall before you leave
How do you solve a Einstein Photoelectric Equation question? A worked example
Pattern: P.PHY.U17.PHOTOELECTRIC_THRESHOLD — Einstein's equation used to find work function, threshold or stopping potential.
- 1
Given
Threshold wavelength of a metal surface, λ₀ = 5.00 × 10⁻⁷ m.
Incident radiation of wavelength λ = 3.00 × 10⁻⁷ m.
Constants: h = 6.63 × 10⁻³⁴ J·s, c = 3.00 × 10⁸ m/s, e = 1.60 × 10⁻¹⁹ C. - 2
Required
The work function W of the metal in eV, and the stopping potential V₀ for the incident radiation.
- 3
Concept
The threshold wavelength marks the point where an absorbed photon has exactly enough energy to free an electron and none left over — so W equals the energy of a photon of wavelength λ₀. For the shorter incident wavelength the photon carries more energy; the surplus above W appears as K_max, which the stopping potential measures through eV₀ = K_max.
- 4
Formula
W = hc/λ₀
K_max = hc/λ − W
eV₀ = K_max - 5
Substitution
W = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (5.00 × 10⁻⁷)
K_max = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (3.00 × 10⁻⁷) − W - 6
Calculation
Numerator hc = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ = 1.989 × 10⁻²⁵ J·m.
W = 1.989 × 10⁻²⁵ / 5.00 × 10⁻⁷ = 3.978 × 10⁻¹⁹ J
In eV: 3.978 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.49 eV
Photon energy at λ: 1.989 × 10⁻²⁵ / 3.00 × 10⁻⁷ = 6.63 × 10⁻¹⁹ J = 4.14 eV
K_max = 4.14 − 2.49 = 1.65 eV = 2.65 × 10⁻¹⁹ J
V₀ = K_max / e = 2.65 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.65 V
Note on constants: h, c and e are physical constants quoted here to three significant figures, and the conversion factor between joules and electronvolts is numerically the electronic charge — none of these is an exact counting number, so the three-figure precision of the data governs the answer. The "1" in the single-photon absorption assumption is exact and does not enter the precision count. - 7
Final answer
W = 2.49 eV; V₀ = 1.66 V.
- 8
Common trap
Reading λ₀ = 5.00 × 10⁻⁷ m as merely a signal that emission will occur, then computing V₀ from the incident photon energy alone: 4.14 eV giving V₀ = 4.14 V. That number appears in option lists. The threshold wavelength is not a yes/no flag — it is the work function in disguise, and it must be converted and subtracted. A second slip in the same problem is mixing units: computing photon energy in joules (6.63 × 10⁻¹⁹) and subtracting a work function quoted in eV (2.49), which yields a number that is neither.
- 9
Similar NEET-style question
The stopping potential for a certain metal is 1.20 V when illuminated by light of wavelength 4.00 × 10⁻⁷ m. Find the threshold wavelength of the metal. *(Work backwards: hc/λ = W + eV₀, solve for W, then λ₀ = hc/W.)*
What to remember before solving Einstein Photoelectric Equation questions
Which Einstein Photoelectric Equation formulas do you need for NEET?
1 formula — click to collapse
Einstein's photoelectric equation
Maximum kinetic energy of photoelectron equals photon energy minus work function. Stopping potential V0 satisfies eV0 = K_max.
| Symbol | Quantity | SI Unit |
|---|---|---|
| K_max | max KE | J |
| h | Planck constant | J*s |
| nu | photon freq | Hz |
| W | work function | J |
| V0 | stopping potential | V |
| e | electron charge | C |
Valid when
- nu > nu_threshold = W/h (else no emission)
- One-photon absorption
Where do students lose marks on Einstein Photoelectric Equation?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Similar Terms
Student writes K_max = hν (forgets W). For frequencies above threshold, K_max = hν − W; W is non-zero.
When it triggers
Photoelectric question with given metal work function or threshold frequency.
How to avoid
Always: K_max = hν − W. If asked for stopping potential V_0: eV_0 = K_max = hν − W.
Root cause: formula misuse
Correction
Always K_max = hν - W. Above threshold (ν > ν_0 = W/h), photons can eject electrons; their max KE equals photon energy minus W (energy needed to free them).
More in Dual Nature of Matter and Radiation: 2 formulas · 1 question pattern from its other lessons.
Einstein Photoelectric Equation questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
All 11 past-paper questions from Dual Nature of Matter and Radiation →
How does NEET ask about Einstein Photoelectric Equation?
1 recurring pattern from past papers — click to collapse
Einstein eq: K_max = hν - W. Find threshold, work function, or stopping potential.
Multi StepMedium
Common distractors
forgets work function
Uses K_max = hν directly
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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