Physics · Dual Nature Of Radiation And Matter · NEET
No. Kmax = hv - phi0 has no intensity term in it. If you make the light brighter (more intense) but keep the same colour (same frequency), each photon still carries the same energy hv, so each electron still gets the same maximum push. Brighter light only sends MORE photons per second, so MORE electrons come out (higher current). It does not make any single electron faster. This is one of the most tested NEET facts.
Kmax = hv - phi0, where h is Planck's constant, v is the frequency of the incident light, and phi0 is the work function of the metal. It can also be written using wavelength as Kmax = hc/lambda - phi0. Because the stopping potential V0 just stops the fastest electron, Kmax = eV0 as well. All three forms describe the same quantity.
The stopping potential V0 is the reverse voltage that is just strong enough to turn back even the fastest photoelectron. To stop that electron, the electric field must do work equal to its kinetic energy. Work done on charge e across potential V0 is eV0. Setting this equal to the fastest electron's energy gives eV0 = Kmax. So measuring V0 in an experiment directly gives you Kmax.
Both, because they are linked by c = v times lambda. Kmax rises linearly with frequency v (Kmax = hv - phi0). In terms of wavelength, Kmax = hc/lambda - phi0, so shorter wavelength (which means higher frequency) gives larger Kmax. NEET may phrase it either way, so be ready to convert using v = c/lambda.
Kmax does NOT simply double. Kmax = hv - phi0 is a straight line, not a proportional relation. If Kmax = hv - phi0 at frequency v, then at frequency 2v it becomes 2hv - phi0, which is more than double the old Kmax but less than 2 times (Kmax + hv). Only the hv part doubles; the work function phi0 stays fixed. Always plug into the equation, do not just scale.
No. If the photon energy hv is less than the work function phi0, then no electron is emitted at all, so Kmax has no meaning (it is not a negative number, it is simply zero emission). Photoemission only starts when hv is at least phi0, that is when frequency reaches the threshold frequency v0. At exactly v0, Kmax = 0.
The maximum kinetic energy of the emitted photoelectrons in the photoelectric effect is independent of:
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:
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the largest amount of movement energy that an electron kicked out by light can have. Electrons deep inside the metal lose some energy escaping, so they come out slower; the ones right at the surface lose the least and come out fastest. That fastest electron's energy is Kmax = hv - phi0.
Use the shortcut: photon energy in eV = 1240 / (wavelength in nm). Then Kmax = (1240/lambda_nm) - phi0(eV). For example, light of 300 nm on a metal of work function 2 eV gives photon energy 1240/300 = 4.13 eV, so Kmax = 4.13 - 2 = 2.13 eV.
Kmax is an energy (measured in joules or eV). Stopping potential V0 is a voltage (in volts) that just stops the fastest electron. They are linked by Kmax = eV0. If you know V0 in volts, Kmax in eV is numerically the same number, because e times V0 gives energy eV0.
Emitted electrons have a range of energies, from nearly zero up to a maximum. The word 'maximum' refers to the surface electrons that escape most easily. In experiments, the stopping potential is set to turn back even these fastest ones, so it measures Kmax, the top of the range.
No. Under the same frequency of light, a metal with a smaller work function phi0 gives a larger Kmax, because Kmax = hv - phi0. For example caesium (low phi0) gives faster electrons than sodium (higher phi0) for the same incident light.
