Bragg reflection and de Broglie wavelength

Explanation

Relationship between Electron Acceleration and de Broglie Wavelength

The kinetic energy gained by an electron accelerated by a potential difference $V$V is $eV$eV. If the mass of the electron is $m$m and the magnitude of its momentum is $p$p, then the kinetic energy can be written as $\frac{p^2}{2m}$2mp2​. Therefore, from the law of conservation of energy, the following relationship holds:

$$eV = \frac{p^2}{2m} \quad \Longrightarrow \quad p = \sqrt{2meV}$$eV=2mp2​⟹p=2meV​

Therefore, the de Broglie wavelength $\lambda$λ of the electron in this case can be expressed as follows:

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$$λ=ph​=2meV​h​

From this equation, we can see that the de Broglie wavelength $\lambda$λ is inversely proportional to the square root of the accelerating voltage ($\lambda \propto \frac{1}{\sqrt{V}}$λ∝V​1​).

First Bragg Reflection (Potential Difference $V$V)

In the first measurement, constructive interference occurs with a potential difference $V$V (wavelength $\lambda$λ), reflection order $n=1$n=1, and $\sin \theta = \frac{1}{8}$sinθ=81​. Substitute this into Bragg's condition.

$$2d \cdot \frac{1}{8} = 1 \cdot \lambda \quad \Longrightarrow \quad \lambda = \frac{d}{4}$$2d⋅81​=1⋅λ⟹λ=4d​

Second Bragg Reflection (Potential Difference $V'$V′)

In the second measurement, the de Broglie wavelength of the electron changes to $\lambda'$λ′ due to the change in potential difference $V'$V′. The incident angle is fixed at $\sin \theta = \frac{1}{2}$sinθ=21​, and constructive interference occurs again at $n=1$n=1.

$$2d \cdot \frac{1}{2} = 1 \cdot \lambda' \quad \Longrightarrow \quad \lambda' = d$$2d⋅21​=1⋅λ′⟹λ′=d

Calculation of the Ratio of Wavelengths and Potential Differences

Calculate the ratio of wavelengths $\frac{\lambda'}{\lambda}$λλ′​ in the two states.

$$\frac{\lambda'}{\lambda} = \frac{d}{\frac{d}{4}} = 4$$λλ′​=4d​d​=4

Here, from the relationship between wavelength and potential difference $\lambda = \frac{h}{\sqrt{2meV}}$λ=2meV​h​, the ratio of wavelengths can be expressed using the ratio of potential differences as follows:

$$\frac{\lambda'}{\lambda} = \frac{\frac{h}{\sqrt{2meV'}}}{\frac{h}{\sqrt{2meV}}} = \sqrt{\frac{V}{V'}}$$λλ′​=2meV​h​2meV′​h​​=V′V​​

Squaring both sides and rearranging:

$$\left(\frac{\lambda'}{\lambda}\right)^2 = \frac{V}{V'} \quad \Longrightarrow \quad 4^2 = \frac{V}{V'} \quad \Longrightarrow \quad \frac{V}{V'} = 16$$(λλ′​)2=V′V​⟹42=V′V​⟹V′V​=16

Since the value we want is $\frac{V'}{V}$VV′​, we take the reciprocal.

$$\frac{V'}{V} = \frac{1}{16}$$VV′​=161​

This result does not depend on specific numerical values ​​such as the electron mass $m$m, electric charge $e$e, or crystal lattice spacing $d$d.

Determining the Input Number

The obtained ratio is an irreducible fraction $\frac{p}{q} = \frac{1}{16}$qp​=161​. The numerator $p=1$p=1 and the denominator $q=16$q=16 give their product $1 \times 16 = 16$1×16=16.

The natural number to input is 16.