If a strong diffraction peak is observed when electrons are incident at an angle i from the normal t

Q: Why do we use $2d \sin \theta$ for the path difference?

When a wave penetrates to the second layer of atoms, it travels an extra distance of $d \sin \theta$ on the way in, and another $d \sin \theta$ on the way out, making the total extra path $2d \sin \theta$.

Q: Why is the glancing angle $90^\circ - i$?

The normal line is, by definition, perpendicular ($90^\circ$) to the crystal plane. Since the incident ray makes an angle $i$ with the normal, the remaining angle down to the plane is $90^\circ - i$.

Q: What exactly does the integer $n$ represent here?

The integer $n$ is the 'order' of diffraction. It represents how many full wavelengths of difference exist between the paths of waves scattering off adjacent planes.

Q: Why are electrons diffracting at all? Aren't they particles?

According to de-Broglie's hypothesis, all moving particles exhibit wave-like behavior. For electrons, this wavelength is comparable to the atomic spacing in crystals, allowing them to diffract just like light!

Q: Would this formula change if the angle $i$ was measured from the plane instead?

Absolutely! If $i$ was the angle with the plane, then $i$ would be the glancing angle $\theta$, and the formula would simply be $2d \sin i = n\lambda$.

Question

If a strong diffraction peak is observed when electrons are incident at an angle $i$ from the normal to the crystal planes with distance $d$ between them (see figure), de-Broglie wavelength $\lambda_{dB}$ of electrons can be calculated by the relationship ($n$ is an integer)

Options:

(A) $d \sin i = n\lambda_{dB}$

(B) $2d \cos i = n\lambda_{dB}$

(C) $2d \sin i = n\lambda_{dB}$

(D) $d \cos i = n\lambda_{dB}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a cross-section of a crystal. The atoms are arranged in neat, parallel layers called **crystal planes**, separated by a constant distance `$d$`.

Now, picture a beam of electrons firing at this crystal. These aren't just particles; thanks to quantum mechanics, they behave like waves with a specific **de-Broglie wavelength**, `$\lambda_{dB}$`.

The electrons strike the crystal at an angle `$i$` measured from the **normal** (the perpendicular line to the surface). They then scatter off at that exact same angle. Our goal is to find the condition where these scattered electron waves combine to form a strong, bright signal—a **diffraction peak**.

### The Master Equation

To get a strong peak, the waves bouncing off the top plane and the waves bouncing off the plane just below it must be perfectly in sync. This happens when the extra distance traveled by the deeper wave—the **path difference**—is an exact integer multiple of the wavelength.

This principle is captured by the famous **Bragg's Law**:

`$\Delta x = n\lambda_{dB}$`

Geometrically, the path difference `$\Delta x$` is related to the spacing `$d$` and the **glancing angle** `$	heta$` (the angle between the incoming beam and the crystal plane itself) by the formula:

`$\Delta x = 2d \sin	heta$`

### Final Calculation

Here is where the trap lies! The problem gives us the angle `$i$` with the normal, not the glancing angle `$	heta$`.

If you look at the geometry, the normal and the crystal plane form a `$90^\circ$` angle. Therefore, the glancing angle is simply the complement of the angle of incidence:

`$	heta = 90^\circ - i$`

Let's substitute this into our path difference equation:

`$\Delta x = 2d \sin(90^\circ - i)$`

Using basic trigonometry, we know that `$\sin(90^\circ - i)$` is exactly equal to `$\cos i$`.

`$\Delta x = 2d \cos i$`

Finally, we equate this path difference to `$n\lambda_{dB}$` to satisfy the condition for constructive interference:

`$2d \cos i = n\lambda_{dB}$`

And there we have it! The correct relationship for a strong diffraction peak in this specific geometry.

Select Answer:

Visualized Solution

$Visual Anchor$

$Bragg’s Law$

$Path Difference$

$Angle Substitution$

$Final Condition$

$The Way Forward$

The Sigma Insight: Matter Waves and de Broglie Relation

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visual Anchor

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bragg’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward