De Broglie Wavelength of an Accelerated Electron

Question

An electron starts from rest and is accelerated through a potential difference of $150\,\text{V}$ . What is its de Broglie wavelength? (Use $hc \approx 1240\,\text{eV}\cdot\text{nm}$ and $m_ec^2 \approx 5.11\times 10^{5}\,\text{eV}$ .)

a. $0.0024\,\text{nm}$
b. $0.10\,\text{nm}$
c. $0.14\,\text{nm}$
d. $8.3\,\text{nm}$

Correct answer

B. $0.10\,\text{nm}$

Full reasoning

1Accelerating through $150\,\text{V}$ gives the electron kinetic energy $K = eV = 150\,\text{eV}$ , far below $m_ec^2 = 5.11\times 10^{5}\,\text{eV}$ , so the nonrelativistic relation $p = \sqrt{2mK}$ is safe.
2Multiply that relation by $c$ to keep everything in electronvolts: $pc = \sqrt{2m_ec^2K} = \sqrt{2(5.11\times 10^{5}\,\text{eV})(150\,\text{eV})}$ .
3Evaluate the product: $2(5.11\times 10^{5})(150) = 1.533\times 10^{8}\,\text{eV}^2$ , so $pc = 1.238\times 10^{4}\,\text{eV}$ .
4Divide into $hc$ : $\lambda = h/p = hc/(pc) = (1240\,\text{eV}\cdot\text{nm})/(1.238\times 10^{4}\,\text{eV}) = 0.100\,\text{nm}$ .
5Sanity-check the scale: $0.1\,\text{nm}$ is one angstrom, comparable to atomic spacing in a crystal, which is exactly the regime in which electron diffraction patterns appear.

Why each choice is right or wrong

Choice A

This is $hc/(m_ec^2) = 1240/(5.11\times 10^{5})$ , the electron's Compton wavelength. It uses the rest energy in place of the momentum, so the accelerating voltage never enters the calculation at all.

Choice B

Correct. $pc = \sqrt{2m_ec^2K} = \sqrt{2(5.11\times 10^{5})(150)} \approx 1.24\times 10^{4}\,\text{eV}$ , so $\lambda = hc/pc = 1240/(1.24\times 10^{4}) \approx 0.10\,\text{nm}$ — about one atomic spacing, which is why electron diffraction from crystals works.

Choice C

This drops the factor of 2 in $p = \sqrt{2mK}$ and uses $pc = \sqrt{m_ec^2K} \approx 8.76\times 10^{3}\,\text{eV}$ . Losing that 2 inflates the wavelength by $\sqrt{2}$ , which is the single most common slip in this calculation.

Choice D

This applies the photon relation $\lambda = hc/E$ with $E = 150\,\text{eV}$ , treating the electron as a massless quantum. For a massive particle the wavelength comes from momentum, and $pc$ is far larger than $K$ when $K \ll mc^2$ .

Related formula

De Broglie wavelength of a slow massive particle

\lambda = \frac{h}{p}, \quad p = \sqrt{2mK}

Assumptions: Nonrelativistic: K must be small compared with the rest energy mc^2. The particle has mass; a photon's wavelength comes from E = hc/lambda instead.

Electronvolt form of the momentum

pc = \sqrt{2(mc^2)K}, \quad \lambda = \frac{hc}{pc}

Assumptions: Working in eV and eV*nm avoids converting to SI units mid-problem. K = e times the accelerating potential difference only if the particle starts from rest and loses no energy on the way.

Sources

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

University Physics Volume 3, Section 6.5: De Broglie's Matter Waves — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
University Physics Volume 3, Section 6.2: Photoelectric Effect — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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