Dual Nature of Radiation and Matter Formula Sheet, Class 12

This dual nature of radiation and matter formula sheet gathers the equations of the Class 12 chapter, built around Einstein's photoelectric equation $K_{max} = h\nu - \phi_0$ . Light behaves as a stream of photons when it knocks electrons out of a metal, and moving particles such as electrons behave as waves with a de Broglie wavelength. Those two ideas give almost every formula below.

The chapter is short but scores well, since most questions are direct substitutions. Keep the constants handy: h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s, e = 1.602 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg, and the shortcut hc ≈ 1240 eV nm. The rest of the syllabus is in the Class 12 chapter tables and the formula sheets hub.

Quick facts

Main equation
$K_{max} = h\nu - \phi_0$
Photon energy
$E = h\nu = \frac{hc}{\lambda}$
Matter waves
$\lambda = \frac{h}{p}$
Handy constant
hc ≈ 1240 eV nm
Energy unit
1 eV = 1.602 × 10⁻¹⁹ J
Courses
Class 12, NEET, JEE Main, AP Physics 2

Photon formulas

Energy and momentum of a photon
Formula	Gives you	Unit	Read more
$E = h\nu$	energy of one photon from its frequency	J	Energy
$E = \frac{hc}{\lambda}$	photon energy from wavelength	J	Wavelength
$E\,(\text{eV}) = \frac{1240}{\lambda\,(\text{nm})}$	quick photon energy in electronvolts	eV	Modern physics
$p = \frac{h}{\lambda} = \frac{E}{c}$	momentum carried by a photon	kg m/s	Momentum
$n = \frac{P}{E} = \frac{P\lambda}{hc}$	photons emitted per second by a source of power P	s⁻¹	Speed of light

A photon has no rest mass but does carry energy and momentum. Brighter light of the same colour means more photons per second, not more energy per photon.

Photoelectric effect formulas

Einstein's photoelectric equation and its relatives
Formula	Gives you	Unit	Read more
$K_{max} = h\nu - \phi_0$	fastest electron energy	J or eV	Work function
$\phi_0 = h\nu_0$	work function from threshold frequency	J or eV	Work function
$\lambda_0 = \frac{hc}{\phi_0}$	longest wavelength that can eject electrons	m	Work function
$eV_0 = K_{max}$	stopping potential	V	Voltage
$V_0 = \frac{h}{e}\nu - \frac{\phi_0}{e}$	straight line of V₀ against ν, slope h/e	V	Modern physics

Below the threshold frequency no electrons come out, however bright the light. Above it, the photocurrent grows with intensity, but the stopping potential depends only on the frequency and the metal.

Example 1: violet light on a metal

Light of wavelength 400 nm falls on a metal with work function 2.0 eV. Find the maximum kinetic energy of the photoelectrons and the stopping potential.

Photon energy: $E = \frac{1240}{400} = 3.10\ \text{eV}$ .

$K_{max} = 3.10 - 2.0 = 1.10\ \text{eV} = 1.10 \times 1.602 \times 10^{-19} = 1.76 \times 10^{-19}\ \text{J}$ .

Stopping potential: $V_0 = \frac{K_{max}}{e} = 1.10\ \text{V}$ . Answer: about 1.1 eV and 1.1 V.

De Broglie wavelength formulas

Wavelength of a moving particle
Formula	Gives you	Unit	Read more
$\lambda = \frac{h}{mv}$	matter wavelength from mass and speed	m	Wavelength
$\lambda = \frac{h}{\sqrt{2mK}}$	matter wavelength from kinetic energy	m	Kinetic energy
$\lambda = \frac{h}{\sqrt{2meV}}$	electron or ion after falling through a potential difference V	m	Voltage
$\lambda_e = \frac{1.227}{\sqrt{V}}\ \text{nm}$	electron shortcut, V in volts	nm	Modern physics

Example 2: an electron through 100 V (exam style)

An electron is accelerated from rest through a potential difference of 100 V. Find its de Broglie wavelength.

Momentum route: $p = \sqrt{2meV} = \sqrt{2(9.11 \times 10^{-31})(1.602 \times 10^{-19})(100)} = 5.40 \times 10^{-24}\ \text{kg m/s}$ .

$\lambda = \frac{6.626 \times 10^{-34}}{5.40 \times 10^{-24}} = 1.23 \times 10^{-10}\ \text{m}$ .

Shortcut check: $1.227/\sqrt{100} = 0.123\ \text{nm}$ . Answer: about 0.123 nm, similar to the spacing of atoms in a crystal, which is why electrons diffract from crystals.

How to use this dual nature of radiation and matter formula sheet in exams

Decide whether the question is about light hitting a metal (photoelectric) or about a particle behaving as a wave (de Broglie).
Work in electronvolts when energies are a few eV; it saves powers of ten.
Convert nm to m only when you use h and c in SI units.
For graphs, remember: K_max against ν is a line of slope h cutting the ν axis at ν₀; photocurrent rises in proportion to intensity.

Give an electron and a proton equal kinetic energies and the proton, being heavier, ends up with the shorter wavelength, since λ varies as 1/√m.

Common mistakes in this chapter

Mixing eV and J. Subtracting a work function in eV from a photon energy in joules gives nonsense. Pick one unit.
Thinking intensity raises K_max. It raises the number of electrons, not their top energy.
Using the electron shortcut for protons. 1.227/√V nm is for electrons only.
Forgetting the square root. Doubling the accelerating voltage cuts λ by a factor of √2, not 2.

The wider set of atomic and nuclear results is on the modern physics summary, and this chapter also appears on the NEET physics formula sheet.

Frequently asked questions

What is Einstein's photoelectric equation?

K_max = hν − φ₀. The energy of a photon, hν, is used partly to free the electron (the work function φ₀) and the rest appears as the electron's maximum kinetic energy.

What is the formula of de Broglie wavelength?

λ = h/p = h/(mv). For a particle of charge q accelerated through voltage V, λ = h/√(2mqV).

What is threshold frequency?

It is the minimum light frequency able to free electrons from a given metal, ν₀ = φ₀/h. Light below it ejects none, whatever its intensity.

Does the stopping potential depend on intensity?

No. It depends only on the frequency of the light and the work function of the metal. Intensity changes the photocurrent.

Is this formula sheet of dual nature of radiation and matter enough for the exam?

It covers the standard formulas of the chapter, but you also need the experiment, the graphs and the reasoning behind them. Use it for quick revision alongside your textbook.