Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

Calculate the \(\lambda\) of X-rays which give a diffraction angle \(2 \theta=16.80^{\circ}\) for a crystal. (Given interplanar distance \(=\) \(0.200 \mathrm{~nm} ;\) diffraction \(=\) first order; \(\sin 8.40^{\circ}=0.1461\) ) (a) \(58.4 \mathrm{pm}\) (b) \(5.84 \mathrm{pm}\) (c) \(584 \mathrm{pm}\) (d) \(648 \mathrm{pm}\)

Step-by-Step Solution

Answer

The correct answer is (a) 58.4 pm.

Step 1: Understand the Bragg's Law

Bragg's Law is given by the formula \( n\lambda = 2d\sin\theta \), where \( \lambda \) is the wavelength, \( n \) is the order of the diffraction, \( d \) is the interplanar distance, and \( \theta \) is the angle of incidence (half of the given \( 2\theta \)).

Step 2: Plug in the given values

Using Bragg's Law, plug in the values: \( n = 1 \) (first-order diffraction), \( d = 0.200 \text{ nm} = 0.200 \times 10^{-9} \text{ m} \), and \( \sin \theta = 0.1461 \).

Step 3: Solve for the Wavelength \( \lambda \)

Rearrange Bragg's Law to \( \lambda = \frac{2d\sin\theta}{n} \). Substitute the values: \( \lambda = \frac{2 \times 0.200 \times 10^{-9} \text{ m} \times 0.1461}{1} \).

Step 4: Calculate \( \lambda \) using the calculation

Calculate the wavelength as \( \lambda = 0.05844 \times 10^{-9} \text{ m} = 58.44 \text{ pm} \).

Step 5: Select the closest option

Compare the calculated wavelength \( 58.44 \text{ pm} \) with the given options. The closest value is option (a) 58.4 pm.

Key Concepts

X-ray diffraction

X-ray diffraction is a fascinating phenomenon that helps scientists explore the structure of crystals at the atomic level. When X-rays encounter a crystal lattice, they scatter off the crystal planes. If these scattered waves are in phase—a condition that depends on specific geometric criteria dictated by Bragg's Law—they constructively interfere to produce visible diffraction patterns.
Such patterns provide valuable insights into the crystal's internal structure.

Understanding this concept is crucial for fields like material science and chemistry as the patterns reveal how atoms are arranged in a material. The angle at which diffraction occurs is related to the wavelength of the X-rays and the distance between atomic layers in the crystal. This relationship allows researchers to determine unknown crystal structures using known X-ray wavelengths or vice versa. This process is highly accurate and is integral in the fields of biology and pharmaceuticals where understanding molecular structure is important.

wavelength calculation

Calculating the wavelength of X-rays using Bragg's Law involves a straightforward yet crucial equation:

• Bragg's Law states that: \[ n\lambda = 2d\sin\theta \]

In this equation, "\(n\)" is the integer order of diffraction, "\(\lambda\)" represents the X-ray wavelength, "\(d\)" is the interplanar spacing, and "\(\theta\)" is the angle between the incident X-rays and the crystal plane, which is often given as half of the diffraction angle. This is because the full angle, \(2\theta\), is the angle directly measured in X-ray experiments.
Through this process, once you have the interplanar spacing and the diffraction angle, you can easily solve for the wavelength if the order of diffraction is known. This calculation is crucial for matching X-rays to known materials and is an essential procedure in materials research and analysis.

interplanar spacing

Interplanar spacing, commonly denoted as "\(d\)", refers to the distance between parallel planes of atoms in a crystal.
This spacing is a fundamental structural feature of crystals, playing a key role in determining their diffraction patterns.

• Bragg's Law highlights this by implicating "\(d\)" as directly influencing the angle at which X-rays will diffract around a crystal.

Materials with different interplanar spacings can be distinguished based on their unique diffraction patterns resulting from the specific spacing.
By analyzing these patterns, scientists can infer critical details about the crystal's internal configuration and the nature of atomic interactions. This understanding aids significantly in the characterization of new materials and enhances our comprehension of existing ones across various industries.