JEE Mains Physics · Dual Nature of Radiation and Matter

Photon Energy, Momentum and Threshold

Light comes in photons of energy hν = hc/λ and momentum h/λ; a source's power is the number of photons per second times that energy, and a photon frees an electron only if its energy beats the metal's work function.

All Dual Nature of Radiation and Matter notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, eighteen of them multiple choice, and two from 2026. Seven are about photon energy: five count the photons a source emits each second, one places a photon in the spectrum from its energy, and one finds the energy an atom keeps after absorbing one photon and emitting another. Five are about the momentum light carries. Eight compare a photon with the work function: a threshold wavelength or frequency, or which metal, colour or lamp can eject electrons at all.

Concept 1 of 3: Photon energy and photons per second

Light arrives in packets. Each packet, a photon, carries energy hν, so blue light carries more per photon than red. A lamp's power is just the number of photons it sends out each second times the energy of one. So at the same power, a red lamp must send out more photons than a blue one.

Definition

$E = h\nu = \dfrac{hc}{\lambda}$ . In electron-volts, $E = \dfrac{1240}{\lambda\,(\text{nm})}$ eV. Some papers give $hc = 1242$ eV nm; use the value printed.
$1\ \text{eV} = 1.6 \times 10^{-19}$ J.
Photons per second from a source of power P: $n = \dfrac{P}{E} = \dfrac{P\lambda}{hc}$ . At equal power, $n \propto \lambda$ .
A beam of intensity I through an area A: $n = \dfrac{IA\lambda}{hc}$ .
Two sources: $\dfrac{P_1}{P_2} = \dfrac{n_1/\lambda_1}{n_2/\lambda_2}$ .
An atom that absorbs $\lambda_1$ and emits $\lambda_2$ keeps $hc\left(\dfrac{1}{\lambda_1} - \dfrac{1}{\lambda_2}\right)$ .
Band from frequency: radio, microwave, infrared, visible (about $4$ – $7.5 \times 10^{14}$ Hz), ultraviolet, X-rays, gamma rays, in rising order.

Photon energy and photon count

E = h\nu = \frac{hc}{\lambda}, \qquad n = \frac{P}{E} = \frac{P\lambda}{hc}

nphotons emitted per second
Ppower of the source

Worked example

A 4 mW laser emits light of wavelength 495 nm. How many photons does it emit each second? (h = 6.6 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s)

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The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q11Moderate

Example 1 · Dual Nature of Radiation and Matter · Photon Energy, Momentum and Threshold

Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is

\_\_\_\_

. (Given :

h = 6.63 \times10^{- 34}

J.s and

c = 3 \times10^{8}\text{ }m/s

)

At equal power, the longer wavelength sends more photons

Each red photon carries less energy than a blue one, so a red lamp must send out more of them to deliver the same power. Photons per second go as λ, not as 1/λ.

A power ratio is not a photon ratio

Power is the photon count times the energy of one photon, P = nhc/λ. When both the count and the wavelength change, write P₁/P₂ = (n₁/λ₁)/(n₂/λ₂) and solve for the unknown.

Use the hc the paper gives

Both 1240 eV nm and 1242 eV nm appear in JEE papers. The options are sometimes close enough that the wrong constant picks the wrong one.

Drill 6 more on photon energy and photons per second

Concept 2 of 3: Photon momentum and the push of light

A photon has no mass but it does carry momentum, p = h/λ = E/c. When light is absorbed, the surface takes that momentum. When light bounces straight back, the momentum reverses, so the surface takes twice as much. A free body that emits a photon recoils the other way.

Definition

$p = \dfrac{h}{\lambda} = \dfrac{E}{c}$ . A shorter wavelength means more momentum and more energy.
Light of total energy E absorbed: momentum delivered $E/c$ . Reflected straight back: $2E/c$ .
A steady beam of power P: force $P/c$ if absorbed, $2P/c$ if reflected.
A pulse of power P lasting t carries $E = Pt$ .
A mirror hanging on a thread: the pulse gives it momentum $mv = 2E/c$ ; then it swings like a pendulum, and for a small angle $v = \theta\sqrt{gl}$ .
A free body of mass M that emits a photon recoils with momentum $h\nu/c$ , so it also gains kinetic energy $p^{2}/2M$ . Its internal energy falls by the photon's energy plus that recoil energy.

Photon momentum and force of light

p = \frac{h}{\lambda} = \frac{E}{c}, \qquad F_{\text{absorbed}} = \frac{P}{c}, \quad F_{\text{reflected}} = \frac{2P}{c}

Worked example

A light pulse of power 30 W lasts 2 ms and falls normally on a mirror that reflects all of it. Find the momentum given to the mirror and the force on it while the pulse lasts. (c = 3 × 10⁸ m/s)

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The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q5Moderate

Example 2 · Dual Nature of Radiation and Matter · Photon Energy, Momentum and Threshold

A small object at rest, absorbs a light pulse of power

20\text{ }mW

and duration

300\text{ }ns

. Assuming speed of light as

3 \times10^{8}\text{ }m/s

, the momentum of the object becomes equal to :

Reflection doubles the push

Absorbed light delivers E/c; light reflected straight back delivers 2E/c, because its momentum reverses. Read the stem for absorbed or reflected before using either.

Energy and momentum rise together

Both E = hc/λ and p = h/λ grow as the wavelength falls. A statement that a shorter wavelength gives a photon less momentum or less energy is false.

Recoil takes energy too

A free body that emits a photon of energy hν loses more than hν of internal energy. The extra is the recoil kinetic energy p²/2M, with p = hν/c.

Drill 4 more on photon momentum and the push of light

Concept 3 of 3: Work function, threshold frequency and threshold wavelength

An electron is held inside a metal. The least energy that frees one is the work function φ. A photon either has that much energy or it does not, and one photon acts on one electron. So there is a lowest frequency, and a longest wavelength, that can eject anything. Below it, no amount of brightness helps.

Definition

$\phi = h\nu_0 = \dfrac{hc}{\lambda_0}$ . In handy units, $\lambda_0\,(\text{nm}) = \dfrac{1240}{\phi\,(\text{eV})}$ .
Emission happens only if $h\nu > \phi$ , that is $\nu > \nu_0$ or $\lambda < \lambda_0$ .
A source below threshold ejects nothing, whatever its power.
If the stem gives the angular frequency ω, use $\nu = \omega/2\pi$ .
Visible light runs from about 400 nm (violet, 3.1 eV) to 700 nm (red, 1.8 eV). Infrared photons carry less than 1.8 eV.

Threshold

\phi = h\nu_0 = \frac{hc}{\lambda_0}, \qquad \lambda_0\,(\text{nm}) = \frac{1240}{\phi\,(\text{eV})}

Worked example

Light of wavelength 500 nm falls on three metals with work functions 2.0 eV, 2.3 eV and 2.8 eV. Which of them emit electrons? Find the threshold wavelength of each. (hc = 1240 eV nm)

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The same idea in a real exam question:

JEE Mains · 2023 · 31 January 2023 · Q92Moderate

Example 3 · Dual Nature of Radiation and Matter · Photon Energy, Momentum and Threshold

If the two metals

A

and

B

are exposed to radiation of wavelength

350\text{ }nm

. The work functions of metals A and B are 4.8eV and 2.2eV. Then choose the correct option.

Brightness does not lower the threshold

A brighter lamp sends more photons of the same energy. If one photon cannot free an electron, a thousand of them cannot either, so a powerful lamp below threshold still ejects nothing.

Angular frequency is 2πν

When a stem gives ω in rad/s, the photon energy is hω/2π. Using hω makes the energy 6.28 times too large.

Longest wavelength, lowest frequency

The threshold is the LONGEST wavelength that ejects electrons and the LOWEST frequency that does. Shorter wavelengths and higher frequencies work; longer and lower do not.

Drill 7 more on work function, threshold frequency and threshold wavelength

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 (3)

Photon energy and photons per second
Photon energy and photon count
$E = h\nu = \frac{hc}{\lambda}, \qquad n = \frac{P}{E} = \frac{P\lambda}{hc}$
Photon momentum and the push of light
Photon momentum and force of light
$p = \frac{h}{\lambda} = \frac{E}{c}, \qquad F_{\text{absorbed}} = \frac{P}{c}, \quad F_{\text{reflected}} = \frac{2P}{c}$
Work function, threshold frequency and threshold wavelength
Threshold
$\phi = h\nu_0 = \frac{hc}{\lambda_0}, \qquad \lambda_0\,(\text{nm}) = \frac{1240}{\phi\,(\text{eV})}$

Watch out for (9)

At equal power, the longer wavelength sends more photons
→ Photon energy and photons per second
A power ratio is not a photon ratio
→ Photon energy and photons per second
Use the hc the paper gives
→ Photon energy and photons per second
Reflection doubles the push
→ Photon momentum and the push of light
Energy and momentum rise together
→ Photon momentum and the push of light
Recoil takes energy too
→ Photon momentum and the push of light
Brightness does not lower the threshold
→ Work function, threshold frequency and threshold wavelength
Angular frequency is 2πν
→ Work function, threshold frequency and threshold wavelength
Longest wavelength, lowest frequency
→ Work function, threshold frequency and threshold wavelength

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