Skip to main content

Wavelength of photoelectrons from X-rays

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).

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.

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

Given: λ=0.9 A˚=0.9×10−10\lambda = 0.9\ \text{Å} = 0.9\times10^{-10} m, ϕ=2\phi = 2 eV, h=6.626×10−34h = 6.626\times10^{-34} J s, c=3×108c = 3\times10^8 m/s, m=9.1×10−31m = 9.1\times10^{-31} kg, e=1.6×10−19e = 1.6\times10^{-19} C.

Photon energy

E=hcλ=6.626×10−34×3×1080.9×10−10=2.2087×10−15 J=13804 eV\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

KE=E−ϕ=13804−2=13802 eV=13802×1.6×10−19=2.2083×10−15 J\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

p=2m KE=2×9.1×10−31×2.2083×10−15=6.34×10−23 kg m s−1λe=hp=6.626×10−346.34×10−23=1.045×10−11 m\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 λe≈1.05×10−11\lambda_e \approx 1.05\times10^{-11} m =0.105= 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.)

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electrical Engineering Material chapter 1 questions →