MHT-CET Physics · Dual Nature of Radiation and Matter

The de Broglie Wavelength

Every moving particle has a wavelength λ = h/p; for a particle of kinetic energy E this is h/√(2mE), and for an electron accelerated from rest through V volts it is h/√(2meV), about 12.27/√V ångström.

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Why this matters

26 PYQs, 6 of them HARD. Thirteen use λ = h/p with energy — comparing a particle with a photon, changing the kinetic energy, a neutron's wavelength at two temperatures. Nine are electrons accelerated through a potential difference, and four fit the electron's wave round a Bohr orbit. Three cards.

Concept 1 of 3: Wavelength, Momentum and Kinetic Energy

The wavelength depends only on momentum: λ = h/p. For a slow particle p = √(2mE), so λ ∝ 1/√E — doubling the kinetic energy divides λ by √2, and halving λ needs four times the energy, three times more added. A thermal neutron has E ∝ T, so λ ∝ 1/√T. A photon is different: its energy is E = pc, so λ = hc/E. Comparing the two at the same energy or the same wavelength is a common HARD question: for equal wavelengths the particle's kinetic energy is smaller than the photon's by the factor h/(2λmc).

Definition

$\lambda = \dfrac{h}{p} = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mE}}$ , so $\lambda \propto E^{-1/2}$ ; impulse = change in p.
Neutron at temperature T: $\lambda \propto \dfrac{1}{\sqrt{T}}$ (27 °C → 927 °C halves λ).
Photon: $\lambda = \dfrac{hc}{E}$ . Same energy: $\dfrac{\lambda_{\text{photon}}}{\lambda_{e}} = c\sqrt{\dfrac{2m}{E}}$ .
Same wavelength: $\dfrac{KE_e}{E_{\text{photon}}} = \dfrac{h}{2\lambda mc}$ .
Accelerated by a field E from rest: $\lambda = \dfrac{h}{eEt}$ , $\dfrac{d\lambda}{dt} = -\dfrac{h}{eEt^2}$ .

de Broglie wavelength

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}

Worked example

An electron (m = 9.1 × 10⁻³¹ kg) has kinetic energy 100 eV. Its de Broglie wavelength?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q15Hard

Example 1 · Dual Nature of Radiation and Matter · de Broglie Wavelength and Matter Waves

Let

E_{e}

and

E_{p}

represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then

E_{p}/E_{e}

is (speed of electron

= C/100

where C is the velocity of light)

Writing λ ∝ 1/E

Momentum goes as the square root of kinetic energy, so λ ∝ E^(−1/2). Doubling E divides λ by √2, not by 2.

Using λ = h/√(2mE) for a photon

A photon has no rest mass; its wavelength is hc/E. Comparing a photon and a particle means using a different formula for each.

Drill 12 more on wavelength, momentum and kinetic energy

Concept 2 of 3: Electrons Accelerated Through a Potential Difference

An electron that falls through V volts gains kinetic energy eV, so λ = h/√(2meV) ≈ 12.27/√V Å. Everything is a square root: four times the voltage halves the wavelength, and a 50% longer wavelength needs 4/9 of the voltage. For particles of the same charge but different mass, a graph of λ against 1/√V is a straight line of slope h/√(2mq), so the heaviest particle has the shallowest line.

Definition

$\lambda = \dfrac{h}{\sqrt{2meV}} \approx \dfrac{12.27}{\sqrt{V}}$ Å; $\lambda \propto \dfrac{1}{\sqrt{V}}$ .
V → 4V halves λ; V doubled ⇒ λ decreased to $\dfrac{1}{\sqrt{2}}$ times; λ up 50% ⇒ $\dfrac{V_1}{V_2} = \dfrac{9}{4}$ .
From a speed: $V = \dfrac{mv^2}{2e}$ ( $1.6\times10^7$ m/s ⇒ 720 V).
λ–(1/√V) graph: slope $\propto \dfrac{1}{\sqrt{m}}$ ; shallowest line = largest mass.

Accelerated electron

\lambda = \frac{h}{\sqrt{2meV}} \approx \frac{12.27}{\sqrt{V}}\ \text{Å}

Worked example

An electron is accelerated through 100 V. Its de Broglie wavelength?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 1 · Q21Moderate

Example 2 · Dual Nature of Radiation and Matter · de Broglie Wavelength and Matter Waves

An electron accelerated through potential

V_1

has de Broglie wavelength

\lambda

. When potential changes to

V_2

wavelength increases by 50%. The value of

\frac{V_1}{V_2}

Reading 'decreased to 1/√2 times' as 'increased'

A larger voltage gives a larger momentum and a SHORTER wavelength. The options pair the right factor with the wrong direction.

Picking the steepest line as the heaviest particle

On λ against 1/√V, the slope is h/√(2mq): a heavier particle has a SMALLER slope.

Drill 8 more on electrons accelerated through a potential difference

Concept 3 of 3: The Electron's Wave Round a Bohr Orbit

Bohr's condition mvr = nh/2π says the electron's de Broglie wave fits round its orbit a whole number of times: nλ = 2πr. So λ = 2πr/n, and with r ∝ n², λ ∝ n. If the first orbit has radius r, the nth has n²r and the wavelength there is 2πnr. Jumping to a higher orbit therefore lengthens the wavelength.

Definition

$n\lambda = 2\pi r_n$ , so $\lambda_n = \dfrac{2\pi r_n}{n}$ .
With $r_n = n^2 r$ : $\lambda_n = 2\pi n r \propto n$ (4th orbit ⇒ 8πr).

Standing wave on an orbit

n\lambda = 2\pi r_n, \qquad \lambda_n = 2\pi n r_1

Worked example

The first Bohr radius is r. De Broglie wavelength of the electron in the second orbit?

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q29Moderate

Example 3 · Dual Nature of Radiation and Matter · de Broglie Wavelength and Matter Waves

If the radius of the first Bohr orbit is '

r

' then the de-Broglie wavelength of the electron in the 4th orbit will be

Dividing the first-orbit circumference by n

λ = 2πrₙ/n uses the radius of the nth orbit, n²r. With r₁ it gives 2πr/n, which shrinks with n; the correct λ = 2πnr grows.

Drill 3 more on the electron's wave round a bohr orbit

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)

Wavelength, Momentum and Kinetic Energy
de Broglie wavelength
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$
Electrons Accelerated Through a Potential Difference
Accelerated electron
$\lambda = \frac{h}{\sqrt{2meV}} \approx \frac{12.27}{\sqrt{V}}\ \text{Å}$
The Electron's Wave Round a Bohr Orbit
Standing wave on an orbit
$n\lambda = 2\pi r_n, \qquad \lambda_n = 2\pi n r_1$

Watch out for (5)

Writing λ ∝ 1/E
→ Wavelength, Momentum and Kinetic Energy
Using λ = h/√(2mE) for a photon
→ Wavelength, Momentum and Kinetic Energy
Reading 'decreased to 1/√2 times' as 'increased'
→ Electrons Accelerated Through a Potential Difference
Picking the steepest line as the heaviest particle
→ Electrons Accelerated Through a Potential Difference
Dividing the first-orbit circumference by n
→ The Electron's Wave Round a Bohr Orbit

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Related notes

Bohr's model — the orbits these waves fit →