---
title: "Wavelength of photoelectrons from X-rays"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:x-ray-photoelectron-wavelength"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/x-ray-photoelectron-wavelength
site: BCT Engineering
---

# Wavelength of photoelectrons from X-rays

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2076 Asoj, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/x-ray-photoelectron-wavelength
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

X-rays of wavelength 0.9 Å fall on a metal plate having work function of 2 eV. Find the wavelength associated with emitted photoelectrons.

## Short answer

Einstein's equation: kinetic energy = hc/λ - φ. Photon energy hc/λ ≈ 13.8 keV, so KE ≈ 13.8 keV (work function negligible). Non-relativistic de Broglie wavelength λ_e = h/√(2m KE) ≈ 1.05×10⁻¹± m (0.105 Å).

## Answer

The X-ray photon gives its energy to an electron; the electron leaves the metal with kinetic energy $KE = hc/\lambda - \phi$ (Einstein's equation). Its de Broglie wavelength is then $\lambda_e = h/p = h/\sqrt{2m\,KE}$.

**Given:** $\lambda = 0.9\ \text{Å} = 0.9\times10^{-10}$ m, $\phi = 2$ eV, $h = 6.626\times10^{-34}$ J s, $c = 3\times10^8$ m/s, $m = 9.1\times10^{-31}$ kg, $e = 1.6\times10^{-19}$ C.

#### Photon energy

$$
\begin{aligned}
E &= \frac{hc}{\lambda} = \frac{6.626\times10^{-34}\times3\times10^{8}}{0.9\times10^{-10}} \\
&= 2.2087\times10^{-15}\ \text{J} = 13804\ \text{eV}
\end{aligned}
$$

#### Kinetic energy of photoelectron

$$
\begin{aligned}
KE &= E - \phi = 13804 - 2 = 13802\ \text{eV} \\
&= 13802\times1.6\times10^{-19} = 2.2083\times10^{-15}\ \text{J}
\end{aligned}
$$

(The work function is negligible compared with the X-ray energy.)

#### Momentum and wavelength

$$
\begin{aligned}
p &= \sqrt{2m\,KE} = \sqrt{2\times9.1\times10^{-31}\times2.2083\times10^{-15}} \\
&= 6.34\times10^{-23}\ \text{kg m s}^{-1} \\
\lambda_e &= \frac{h}{p} = \frac{6.626\times10^{-34}}{6.34\times10^{-23}} = 1.045\times10^{-11}\ \text{m}
\end{aligned}
$$

**Answer:** wavelength of the photoelectrons $\lambda_e \approx 1.05\times10^{-11}$ m $= 0.105$ Å.

(KE = 13.8 keV is only about 2.7 % of the electron rest energy, 511 keV, so the non-relativistic formula is adequate.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The kinetic energy of the photoelectron is:
   A. hc/lambda + phi
   B. phi - hc/lambda
   C. hc/lambda - phi
   D. hc/lambda x phi
   Answer: C. KE = photon energy minus work function. (Key point: Photoelectric equation)

2. Wavelength of the emitted electron is found from:
   A. lambda = hc/KE
   B. lambda = h/sqrt(2m KE)
   C. lambda = KE/h
   D. The X-ray wavelength directly
   Answer: B. Electron wavelength is h/p with p = sqrt(2m KE). (Key point: de Broglie step)

3. The energy of a 0.9 Å X-ray photon is about:
   A. 13.8 eV
   B. 138 eV
   C. 1.38 MeV
   D. 13.8 keV
   Answer: D. E = 12400/0.9 eV is about 13800 eV. (Key point: Photon energy)

4. The photoelectron wavelength is about:
   A. 0.105 Å
   B. 0.9 Å
   C. 1.05 Å
   D. 10.5 Å
   Answer: A. lambda_e = h/p is about 1.05 x 10^-11 m = 0.105 Å. (Key point: Final answer)

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