Answer to Question #100819 in Quantum Mechanics for Ra

Question

A photon and an electron each hae an energy of 6.0×10^3 eV. To calculate their respective wavelengths. Which of these could be used to probe atomic structures?

Answer 1

The concept that matter behaves like a wave was proposed by Louis de Broglie in 1924. The de Broglie wavelength is

(1) $\lambda=\frac{h}{p}$ ,

where $p$ - the momentum of particle, and $h$ is the Planck constant. Wave-like behavior of matter was first experimentally demonstrated by George Paget Thomson's thin metal diffraction experiment, and independently in the Davisson–Germer experiment both using electrons, and it has also been confirmed for other elementary particles, neutral atoms and even molecules [1]. The rest energy of an electron $E_e=0.511 MeV$, and especially a proton $E_p=0.938 GeV$ , is much larger than the kinetic energy $E_k=6.0 KeV$ specified in the problem; therefore, to estimate the wavelength, one can use the formulas of nonrelativistic mechanics.

$E_k=\frac{mv^2}{2}$ ; $p=mv$

We can rewrite these as

$E_k=\frac{p^2}{2m}$ ; and $p=\sqrt{2mE_k}$

Recalling Einstein's formula for the energy of rest $E=mc^2$ of a body with mass $m$, we find

$p_e=\frac{1}{c}\sqrt{2E_eE_k}$

(2) $p_e=\frac{1}{c} \sqrt{2\cdot 0.511\cdot 10^6\cdot 6\cdot 10^3} \cdot eV =\frac{1}{c}\cdot 7.8\cdot 10^4 eV$

Similarly for a proton we have

$p_p=\frac{1}{c}\sqrt{2E_pE_k}$

(3) $p_p=\frac{1}{c} \sqrt{2\cdot 0.938\cdot 10^9\cdot 6\cdot 10^3} \cdot eV =\frac{1}{c}\cdot 3.4\cdot 10^6 eV$

To find the de Broglie wavelength we should sustitude (2),(3) to (1)

$\lambda_e=\frac{h}{p_e}=\frac{hc}{7.8\cdot 10^4 eV}=\frac{1.24 eV \cdot \mu m}{7.8\cdot 10^4 eV}=0.16\cdot10^{-10}m$ and

$\lambda_p=\frac{h}{p_p}=\frac{hc}{3.4\cdot 10^6 eV}=\frac{1.24 eV \cdot \mu m}{3.4\cdot 10^6 eV}=0.36\cdot10^{-12}m$ .

In the calculations we used the value $hc=1.24 eV \cdot \mu m$ [2]

Answer: $\lambda_e=0.16\cdot10^{-10}m$ ; $\lambda_p=0.36\cdot10^{-12}m$. We can see that for a proton the wavelength is much shorter than that of electron, thus the proton is more usefull to probe atomic structures.

[1] https://en.wikipedia.org/wiki/Matter_wave

[2] https://en.wikipedia.org/wiki/Planck_constant