Physics · Dual Nature Of Radiation And Matter · NEET
The stopping potential is applied as a NEGATIVE voltage on the collector plate (the plate that catches the electrons). A negative plate pushes the negative electrons back. We usually write its magnitude as a positive number V0, but the plate itself is at minus V0. In NEET, if a question asks for the stopping potential 'of A relative to C', the answer often carries a minus sign, like -3 V.
No. Stopping potential does NOT change with intensity. Brighter light (higher intensity) sends out MORE electrons per second, so the saturation current is larger, but each electron still leaves with the same maximum energy. Since V0 measures only the maximum energy (eV0 = Kmax), it stays the same. This is a very common NEET trap.
When the fastest electron (energy Kmax) is just stopped, all its kinetic energy is used up climbing against the electric force. The work done to stop it is charge times voltage = e·V0. So e·V0 = Kmax = (1/2)m·vmax². Here e = 1.6 x 10^-19 C. If you use V0 in volts and e in units of electron charge, then Kmax comes out directly in eV (a neat shortcut).
Work function (phi) is a fixed property of the metal: the minimum energy to pull one electron out. Stopping potential V0 depends on BOTH the metal and the frequency of light: e·V0 = h·nu - phi. Change the light frequency and V0 changes, but phi stays fixed. Work function is about escaping the metal; stopping potential is about the energy left over AFTER escaping.
As you make the collector more negative, slower electrons get turned back first, so the current drops. At V0, even the single fastest electron (with Kmax) is turned back just before reaching the plate. With no electron reaching the collector, the current is zero. Any voltage slightly less negative would let the fastest electrons through, giving a tiny current.
Yes. From eV0 = h·nu - phi, higher frequency light gives higher photon energy, so electrons leave with more energy, and a larger V0 is needed to stop them. A graph of V0 versus frequency nu is a straight line with slope h/e, which is how NEET problems extract Planck's constant.
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:
When two monochromatic lights of frequency nu and nu/2 are incident on a photoelectric metal, their stopping potentials become Vs/2 and Vs respectively. The threshold frequency for this metal is:
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A if the stopping potential of A relative to C is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Volt (V), since it is a potential difference. Its magnitude times the electron charge gives energy in joules, or directly in eV if V0 is in volts.
Yes. If the incident light frequency equals the threshold frequency, the electrons leave with almost zero kinetic energy, so V0 is essentially zero. Below threshold there is no emission at all.
Use eV0 = (1/2)m·vmax². Rearranged, vmax = sqrt(2·e·V0/m), where m = 9.1 x 10^-31 kg is the electron mass.
The slope equals h/e (Planck's constant divided by electron charge). Multiplying the slope by e gives Planck's constant h. The x-intercept gives the threshold frequency.
It is the direct experimental measure of the maximum kinetic energy of photoelectrons. This let physicists confirm Einstein's equation Kmax = h·nu - phi and prove the particle nature of light.
