Physics · Dual Nature Of Radiation And Matter · NEET
h*v is the energy carried by ONE photon (h = Planck's constant 6.63e-34 J.s, v = frequency). phi_0 is the work function, the minimum energy an electron needs to escape the metal surface. Kmax is the maximum kinetic energy of the fastest ejected electron. In simple words: photon energy in, exit fee (phi_0) paid, leftover energy becomes motion. If h*v is less than phi_0, no electron comes out.
In Einstein's photon picture, one electron absorbs exactly one photon. The photon's energy depends only on its frequency (h*v), not on how many photons there are. Intensity means more photons per second, so more electrons are ejected (higher current), but each electron still gets the energy of a single photon. So intensity raises the NUMBER of electrons, not their maximum energy. This is why Kmax depends on frequency and work function only.
Use the shortcut E = 1240 / lambda, where E is in eV and lambda (wavelength) is in nanometres. So photon energy in eV = 1240/lambda(nm). Then Kmax(eV) = 1240/lambda - phi_0(eV). Example: light of 400 nm on a metal with phi_0 = 2.0 eV gives E = 1240/400 = 3.1 eV, so Kmax = 3.1 - 2.0 = 1.1 eV. This saves you from converting joules every time.
h*v is the TOTAL energy delivered by the photon. phi_0 is the FIXED energy cost to leave the metal (a property of the metal only). Kmax is what is LEFT OVER for the electron's speed. Think of a toll road: h*v is the money you carry, phi_0 is the toll, Kmax is the money left after paying. If your money (h*v) is below the toll (phi_0), you cannot even enter, so no emission.
Rewrite the equation as Kmax = h*v - phi_0, which is in the form y = m*x + c with x = frequency, slope m = h (Planck's constant, same for all metals) and intercept c = -phi_0. So plotting Kmax against frequency gives a straight line: its slope gives h, and where it cuts the frequency axis (Kmax = 0) gives the threshold frequency v_0 = phi_0/h.
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 light of frequency 2v_0 (v_0 is the threshold frequency) is incident on a metal plate, the maximum velocity of electrons is v1. When the frequency is increased to 5v_0, the maximum velocity is v2. The ratio v1 : v2 is:
The maximum kinetic energy of the emitted photoelectrons in the photoelectric effect is independent of:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is Kmax = h*v - phi_0. The maximum kinetic energy of a photoelectron equals the photon energy h*v minus the work function phi_0 of the metal. It expresses energy conservation for one photon absorbed by one electron.
Photon energy in eV = 1240 / lambda, where lambda is the wavelength in nanometres. So Kmax(eV) = 1240/lambda(nm) - phi_0(eV). This lets you solve NEET numericals fast without converting to joules.
When Kmax = 0, the photon energy just equals the work function: h*v_0 = phi_0, so v_0 = phi_0/h. Below this threshold frequency v_0, no electrons are emitted no matter how intense or how long the light shines.
He explained three facts that the wave theory could not: Kmax depends on frequency (not intensity), there is a threshold frequency, and emission is instantaneous. His photon idea (light as energy packets h*v) was confirmed by Millikan's experiments, which also gave an accurate value of Planck's constant.
The stopping potential V0 is the voltage that just stops the fastest electron, so e*V0 = Kmax. Therefore Einstein's equation can be written as e*V0 = h*v - phi_0. Measuring V0 for different frequencies lets you find h from the graph slope.
