xSkill: notes, lectures and practice for CBSE and CSE

1.1 The photon and the photoelectric equation Light of frequency nu carries energy in packets called photons, each of energy h nu. When such a photon strikes a metal surface it can eject an electron, provided its energy is at least the work function of the metal. The extra energy appears as the kinetic energy of the electron. Einstein's equation for the photoelectric effect combines these three quantities. The maximum kinetic energy of the emitted electron equals the photon energy minus the work function. The stopping potential is the smallest retarding voltage that just stops the fastest electrons, so it measures the maximum kinetic energy in electron volts.

E = h ν = \frac{h c}{λ}, ϕ = h ν_{0}

Photon energy and the work function

K_{m a x} = h ν - ϕ = e V_{0}

Einstein's photoelectric equation with the stopping potential

1.2 The experimental laws The photoelectric current at saturation is proportional to the intensity of the incident light, because intensity fixes the number of photons arriving per second. The stopping potential does not depend on intensity; it depends on the frequency of the light and on the metal. For frequencies below the threshold value no emission occurs, however intense the beam. The maximum kinetic energy increases linearly with frequency, and the slope of that line is Planck's constant. This graph is the standard experimental confirmation of Einstein's equation.

K_{m a x} \propto ν, slope = h, intercept = - ϕ

Straight line graph of maximum kinetic energy against frequency

1.3 Matter waves De Broglie proposed that particles also have a wave character, with a wavelength inversely proportional to their momentum. The relation is the same one that connects the photon energy and momentum. For an electron accelerated through a potential difference V, the de Broglie wavelength follows from its kinetic energy and can be written in terms of V. The Davisson and Germer experiment scattered electrons from a nickel crystal and observed diffraction maxima, which confirmed the wave nature of electrons. The same idea explains the discrete energy levels of atoms. Keep both pictures ready, because a question can ask for the wave argument and the photon argument in the same answer.

λ = \frac{h}{p} = \frac{h}{m v}

de Broglie wavelength of a particle

λ = \frac{h}{2 m K} = \frac{h}{2 m e V}

de Broglie wavelength for a charged particle accelerated through V

Exam

Write hc as 1240 eV nanometre, or hc equal to 1242 eV nanometre as used in the NCERT examples, to convert quickly between wavelength and photon energy in electron volts.