Chemistry · Structure Of Atom · NEET
When light hits a metal, electrons come out with different speeds. The fastest ones have the maximum kinetic energy (KEmax). Now you connect a battery the reverse way, so the plate that collects electrons becomes negative and pushes electrons back. As you raise this reverse voltage, slower electrons stop first. The stopping potential V0 is the exact reverse voltage at which even the fastest electron is stopped and the current becomes zero. So V0 is a way to measure KEmax without a stopwatch.
The work done by the reverse voltage to stop the fastest electron equals its kinetic energy. Work = charge × voltage = e × V0. So e × V0 = KEmax. Here e = 1.6 × 10^-19 C. Combined with Einstein's equation, e × V0 = h(nu) - h(nu0) = h(nu) - W0. This single line answers most NEET questions on this topic.
No. This is the most important point. Brighter light means MORE photons, so MORE electrons come out and the current is bigger. But each photon still gives the same energy h(nu) to one electron, so KEmax does not change. Since eV0 = KEmax, the stopping potential stays the SAME when you only increase intensity. NEET loves this trap.
Yes. If you use light of higher frequency (nu), each photon carries more energy h(nu). So KEmax = h(nu) - W0 increases, and therefore V0 increases. In short: intensity changes current only; frequency changes stopping potential. Remember: 'Frequency feeds the stopping potential.'
Work function (W0 = h(nu0)) is a fixed property of the metal - the minimum energy needed just to pull an electron out, at the surface, with zero leftover speed. Stopping potential is about the electron AFTER it comes out - it measures the extra kinetic energy the fastest electron carries. Work function is constant for a metal; stopping potential grows as you raise the light frequency.
The collector plate is made negative on purpose so it repels the incoming electrons. This reverse (retarding) field does negative work on the electrons and slows them down. At voltage V0 the field has removed all the kinetic energy of the fastest electron, so it just fails to reach the plate and the current becomes zero. That is why V0 is a retarding/negative potential.
Plot V0 on the y-axis and frequency (nu) on the x-axis. From eV0 = h(nu) - W0, you get V0 = (h/e)(nu) - (W0/e). This is a straight line. The slope is h/e (same for every metal - a NEET favourite), it cuts the x-axis at the threshold frequency nu0, and the negative y-intercept is -W0/e. Different metals give parallel lines (same slope, different intercepts).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Volt (V), because it is a potential difference. When multiplied by the electron charge e (in coulombs), eV0 gives energy in joules.
Yes. If the light frequency equals the threshold frequency (nu = nu0), the electron comes out with zero kinetic energy, so KEmax = 0 and V0 = 0. Below nu0 no electron comes out at all.
Yes. The slope is always h/e (Planck's constant divided by electron charge), which is a universal constant. Only the intercept (which depends on work function) changes from metal to metal.
No. Distance changes the intensity (how bright the light looks), not the frequency of each photon. So KEmax and stopping potential stay the same.
Use eV0 = h(nu) - W0. Find the photon energy h(nu) (or hc/lambda), subtract the work function W0 to get KEmax, then divide by e = 1.6 × 10^-19 C to get V0 in volts.
