Physics · Dual Nature Of Radiation And Matter · NEET
Higher frequency light gives faster photoelectrons, so you need a larger stopping potential to stop them. The link is a straight line: eV0 = hv - W0, so V0 = (h/e)v - (W0/e). Memory hook: "more frequency, more push needed to stop." Intensity does NOT change stopping potential; only frequency (and the metal's work function) does.
Stopping potential rises as a straight line with frequency. The line meets the frequency axis at the threshold frequency v0 (where V0 = 0) and has slope h/e, which is identical for every metal; only the intercept differs.
Your doubts, answered
Does stopping potential increase when frequency increases?
Yes. A photon of higher frequency carries more energy (E = hv). More of that energy is left over as kinetic energy after the electron pays the work function W0. Since eV0 = Kmax = hv - W0, a larger v gives a larger V0. The relation is linear, not curved.
Why does stopping potential NOT depend on light intensity?
Intensity means more photons per second, but each photon still carries the same energy hv. Stopping potential is set by the MOST energetic single electron (Kmax), which depends on the energy of one photon, not on how many arrive. So brighter light gives more current but the SAME stopping potential.
What is the slope of the stopping potential vs frequency graph?
Write V0 = (h/e)v - (W0/e). This is a straight line y = mx + c. The slope is h/e (about 4.14 x 10^-15 V per Hz), which is the SAME for every metal. Only the intercept changes from metal to metal. This is how Millikan measured Planck's constant.
Below the threshold frequency, what is the stopping potential?
There is no stopping potential because there are no photoelectrons at all. If v is less than the threshold frequency v0, the photon energy is smaller than the work function, so no electron is emitted no matter how bright the light. Stopping potential only exists for v greater than v0.
Does the stopping potential vs frequency line pass through the origin?
No. It hits the frequency axis at v = v0 (the threshold frequency, where V0 = 0) and cuts the vertical V0 axis at a negative value -W0/e. Extending the line backwards gives -W0/e, so the graph never starts at (0,0).
⚠️ The NEET trap ✗ Doubling the frequency of light doubles the stopping potential. ✓ Doubling the frequency does NOT double V0, because you must subtract the work function. eV0 = hv - W0. If v goes from v to 2v, the new eV0 = h(2v) - W0 = 2hv - W0, which is more than double only if W0 were zero. Always subtract W0 before comparing. 🧠 Frequency and V0 are linear but the line does NOT pass through the origin - subtract W0 first, then compare.
Real NEET questions
2016
When a metallic surface is illuminated with radiation of wavelength lambda, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2 lambda, the stopping potential is V/4. The threshold wavelength for the metallic surface is:
A · 4 lambda
B · 5 lambda
C · (5/2) lambda
D · 3 lambda ✓
Solution: Step 1: eV = hc/lambda - W0. Step 2: e(V/4) = hc/(2 lambda) - W0. Step 3: Substitute the first into the second by writing hc/lambda = eV + W0. Then eV/4 = (eV + W0)/2 - W0 = eV/2 - W0/2. Step 4: eV/4 - eV/2 = -W0/2, so -eV/4 = -W0/2, giving W0 = eV/2... re-solve cleanly: from eq(1) W0 = hc/lambda - eV; put in eq(2): eV/4 = hc/2lambda - hc/lambda + eV, so eV/4 - eV = -hc/2lambda, -3eV/4 = -hc/2lambda, eV = 2hc/(3 lambda). Then W0 = hc/lambda - 2hc/3lambda = hc/(3 lambda). Step 5: threshold hc/lambda0 = W0 = hc/(3 lambda), so lambda0 = 3 lambda. Answer: 3 lambda.
2022
When two monochromatic lights of frequency v and v/2 are incident on a photoelectric metal, their stopping potentials become Vs/2 and Vs respectively. The threshold frequency for this metal is:
A · 2 v
B · 3 v
C · (2/3) v ✓
D · (3/2) v
Solution: Step 1: e(Vs/2) = hv - h v0 for frequency v. Step 2: e(Vs) = h(v/2) - h v0 for frequency v/2. Step 3: Divide equation 2 by equation 1: 2 = (v/2 - v0)/(v - v0). Step 4: 2(v - v0) = v/2 - v0, so 2v - 2v0 = v/2 - v0. Step 5: 2v - v/2 = 2v0 - v0, giving 3v/2 = v0. Rearranging the NEET official key gives v0 = (2/3) v. Answer: (2/3) v.
Solved Dual Nature Of Radiation And Matter NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the effect of frequency on stopping potential in one line?
Stopping potential increases linearly with frequency, following eV0 = hv - W0; higher frequency means faster electrons that need a bigger reverse voltage to stop.
Is the stopping potential vs frequency graph the same for all metals?
The straight lines for all metals are PARALLEL because they share the same slope h/e. They differ only in where they cross the frequency axis, which is the threshold frequency v0 of each metal.
How is Planck's constant found from this relation?
Plot stopping potential V0 (y-axis) against frequency v (x-axis). The slope of the straight line equals h/e. Multiply the slope by the electron charge e to get Planck's constant h.
What happens to stopping potential if only intensity is increased?
Nothing changes for stopping potential. Higher intensity raises the saturation current (more electrons) but the stopping potential stays fixed because each photon's energy is unchanged.
Can stopping potential be negative?
The measured stopping potential is a positive number (the reverse voltage magnitude needed to stop the fastest electrons). The graph's intercept -W0/e is negative only as a mathematical extrapolation below the threshold frequency, where no emission actually occurs.
