JEE Mains Chemistry · Structure of Atom

Photons, Planck's Quantum and the Photoelectric Effect

Light comes in packets of energy hν. This page turns wavelength, frequency and wavenumber into photon energy, per photon or per mole, and uses it to eject electrons from a metal.

All Structure of Atom notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, twelve of them numerical-answer, and five from 2026. Eleven convert between wavelength, frequency, wavenumber and photon energy, count photons, or test black-body facts; eight use the photoelectric equation with a work function or threshold frequency. Two ideas cover the page.

Concept 1 of 2: Photon energy from wavelength, frequency or wavenumber

A photon's energy is fixed by its frequency:

E=h\nu

. Wavelength and wavenumber are two other ways of stating the same frequency, through

c=\nu\lambda

and

\bar\nu=1/\lambda

. So shorter wavelength means more energy. To compare photons given in different forms, turn them all into frequency first.

Definition

$c=\nu\lambda$ , $\bar\nu=\dfrac{1}{\lambda}$ .
$E=h\nu=\dfrac{hc}{\lambda}=hc\bar\nu$ , with $h=6.626\times10^{-34}\ \mathrm{J\,s}$ .
With $\bar\nu$ in $\mathrm{cm^{-1}}$ , use $c=3\times10^{10}\ \mathrm{cm\,s^{-1}}$ .
Per mole of photons: multiply by $N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}$ .
Number of photons from a source of power $P$ in time $t$ : $n=\dfrac{Pt}{hc/\lambda}$ .
A black body absorbs and emits all frequencies. Its spectrum depends only on temperature. As $T$ rises, the peak moves to shorter wavelength ( $\lambda_{max}T$ is constant). Planck explained it by quantising energy in units of $h\nu$ .

Photon energy

E=h\nu=\frac{hc}{\lambda}=hc\bar\nu

Worked example

Photon P has wavelength 250 nm, photon Q has frequency

2\times10^{15}\ \mathrm{s^{-1}}

, and photon R has wavenumber

3\times10^{4}\ \mathrm{cm^{-1}}

. Arrange them by energy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q34Moderate

Example 1 · Structure of Atom · Photons, Planck's Quantum and the Photoelectric Effect

The wavelength of photon ' A ' is 400 nm. The frequency of photon ' B ' is

10^{16}{\text{ }s}^{- 1}

. The wave number of photon ' C ' is

10^{4}{\text{ }cm}^{- 1}

. The correct order of energy of these photons is :

Wavenumber in cm⁻¹ needs c in cm s⁻¹

E=hc\bar\nu

only works if the units match. With

\bar\nu

\mathrm{cm^{-1}}

, take

c=3\times10^{10}\ \mathrm{cm\,s^{-1}}

. Using

3\times10^{8}

makes the energy 100 times too small.

Energy ratio is the inverse wavelength ratio

\dfrac{E_1}{E_2}=\dfrac{\lambda_2}{\lambda_1}

. A 900 nm photon has one third the energy of a 300 nm photon, not three times.

Drill 10 more on photon energy from wavelength, frequency or wavenumber

Concept 2 of 2: Photoelectric effect and work function

One photon hits one electron. Part of its energy, the work function

W_0

, pulls the electron out of the metal. The rest becomes the electron's kinetic energy. If the photon has less energy than

W_0

, nothing comes out, however bright the light. Brighter light means more photons, so more electrons, but each electron gets the same energy.

Definition

$h\nu=W_0+\tfrac12mv^2$ , with $W_0=h\nu_0$ .
Threshold wavelength: $\lambda_0=\dfrac{hc}{W_0}$ , the longest wavelength that ejects electrons.
Below $\nu_0$ : no emission at any intensity. Above $\nu_0$ : emission is instant.
Current $\propto$ intensity. Kinetic energy $\propto(\nu-\nu_0)$ , and does not depend on intensity.
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$ . Shortcut: $E\,(\mathrm{eV})=\dfrac{1240}{\lambda\,(\mathrm{nm})}$ .

Einstein's photoelectric equation

h\nu=h\nu_0+\tfrac{1}{2}mv^2

Worked example

A metal has work function 2.0 eV. Light of wavelength 310 nm falls on it. Find the maximum kinetic energy of the electrons and the threshold wavelength.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q43Moderate

Example 2 · Structure of Atom · Photons, Planck's Quantum and the Photoelectric Effect

The work functions of two metals (

M_{A}

and

M_{B}

) are in the

1:2

ratio. When these metals are exposed to photons of energy 6 eV , the kinetic energy of liberated electrons of

M_{A}:M_{B}

is in the ratio of 2.642 : 1. The work functions (in eV ) of

M_{A}

and

M_{B}

are respectively.

Intensity changes the current, not the energy

Brighter light gives more electrons per second. It does not give faster electrons. Only a higher frequency raises the kinetic energy.

Put eV and J on the same footing

Work functions are usually in eV and

h\nu

comes out in J. Convert with

1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}

before you subtract.

Drill 7 more on photoelectric effect and work function

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Photon energy from wavelength, frequency or wavenumber
Photon energy
$E=h\nu=\frac{hc}{\lambda}=hc\bar\nu$
Photoelectric effect and work function
Einstein's photoelectric equation
$h\nu=h\nu_0+\tfrac{1}{2}mv^2$

Watch out for (4)

Wavenumber in cm⁻¹ needs c in cm s⁻¹
→ Photon energy from wavelength, frequency or wavenumber
Energy ratio is the inverse wavelength ratio
→ Photon energy from wavelength, frequency or wavenumber
Intensity changes the current, not the energy
→ Photoelectric effect and work function
Put eV and J on the same footing
→ Photoelectric effect and work function

Test yourself on Structure of Atom

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Structure of Atom chapter test Drill all 19 questions