For crystal diffraction experiments, wavelengths on the order of 0.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; (b) an electron; (c) an alpha particle (m = 6.64 * 10^-27 kg).

According to de Broglie, Light has dual nature as in some situations it behaves like waves and in others like particles. Electrons, which sometimes exhibits particle nature may also behave like wave in some situations.

When a particle acts like a wave, it carries a wavelength which is known as de Broglie wavelength.

Wavelength, λ = 0.20 nm = 2 x 10^-10 m

Charge on electron, e = 1.6 x 10^-19 C

Energy of photon is given by:

$$\begin{gathered} E=hf \\ =\frac{hc}{e\lambda } \\ =\frac{\left(6.626*{10}^{-34}Js\right)\left(3*{10}^{8}m/s\right)}{\left(1.6*{10}^{-19}C\right)\left(2*{10}^{-10}m\right)} \\ =6211.875eV \end{gathered}$$

Therefore, the energy of photon is 6211.875 eV.

Wavelength, λ = 0.20 nm = 2 x 10^-10 m

De Broglie wavelength is given by

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2meE}}$

Where, Plank’s constant, h = 6.626 x 10^-34 Js

Charge on electron, e = 1.6 x 10^-19 C

E is the Energy of the electron in eV

Mass of electron, m_e = 9.11 x 10^-31 Kg

Energy of electron is given by:

role="math" localid="1664007101723" $$\begin{gathered} E=\frac{h^2}{2me{\lambda }^2} \\ E=\frac{h^2}{2m_{e}e{\lambda }^2} \\ =\frac{\left(6.626*{10}^{-34}Js\right)}{2\left(9.11*{10}^{-31}Kg\right)\left(1.6*{10}^{-19}C\right){\left(2*{10}^{-10}\right)}^2} \\ =37.65eV \end{gathered}$$

Therefore, the energy of electron is 37.65 eV.

Wavelength, λ = 0.20 nm = 2 x 10^-10 m

Mass of alpha-particle, m_a = 6.64 x 10^-27 Kg

Energy of alpha-particle is given by:

$$\begin{gathered} E=\frac{h^2}{2m_{e}e{\lambda }^2} \\ =\frac{\left(6.626*{10}^{-34}Js\right)}{2\left(6.64*{10}^{-27}Kg\right)\left(1.6*{10}^{-19}C\right){\left(2*{10}^{-10}\right)}^2} \\ =0.0052eV \end{gathered}$$

Therefore, the energy of alpha-particle is 0.0052 eV.