Question

Under appropriate conditions, molybdenum emits Xrays that have a characteristic wavelength of \(0.711 \AA .\) These X-rays are used in diffraction experiments to determine the structures of molecules. (a) Why are X-rays, and not visible light, suitable for the determination of structure at the atomic level? (b) How fast would an electron have to be moving to have the same wavelength as these X-rays?

Short answer

(a) X-rays are suitable for determining atomic-level structures due to their shorter wavelengths, which allow them to probe matter more effectively than visible light. This is because of Bragg's Law, which requires the diffraction grating spacing to be comparable to the wavelength of the incident radiation. Since atomic and molecular dimensions are typically in the range of angstroms, X-rays with wavelengths of the same order can accurately measure structures. (b) To calculate the speed of an electron with the same wavelength as these X-rays, we use the de Broglie's wavelength equation: \(\lambda = \frac{h}{mv}\). The given wavelength is \(\lambda = 7.11 \times 10^{-11} m\). Substituting the values, we get the electron speed as \(v = 1.30 \times 10^6 m/s\).

Step-by-step solution

Properties of X-rays

X-rays have shorter wavelengths which allow them to probe the atomic-level structure of matter more effectively than visible light. This is due to the Bragg's Law, which requires the diffraction grating spacing to be comparable to the wavelength of the incident radiation. Since the dimensions of atoms and molecules are typically in the range of angstroms (\(\overset{\circ}{A}\)), X-rays with wavelengths of the same order can be used to accurately measure their structures.

Molybdenum X-ray wavelength

We are given the characteristic wavelength of X-rays emitted by molybdenum as \(0.711 \overset{\circ}{A}\). We need to convert this value to the appropriate SI unit, meters, to perform further calculations. One angstrom is equal to \(10^{-10}\) meters. Therefore, the wavelength in meters is given by: $$ \lambda = 0.711 \overset{\circ}{A} \times 10^{-10} \frac{m}{\overset{\circ}{A}} = 7.11 \times 10^{-11} m $$

De Broglie's wavelength equation

The de Broglie's wavelength equation relates the wavelength of a particle (e.g., an electron) to its momentum: $$ \lambda = \frac{h}{p} $$ where \(\lambda\) is the wavelength, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} Js\)), and \(p\) is the momentum of the particle. For an electron, its momentum is given by: $$ p = mv $$ where \(m\) is the mass of the electron (\(9.11 \times 10^{-31} kg\)) and \(v\) is its speed.

Calculating the electron speed

Using the de Broglie's wavelength equation, we can calculate the electron speed: $$ \lambda = \frac{h}{mv} $$ Solve for \(v\): $$ v = \frac{h}{m \lambda} $$ Substitute the values: $$ v = \frac{6.626 \times 10^{-34} Js}{(9.11 \times 10^{-31} kg)(7.11 \times 10^{-11} m)} $$ Finally, calculate the speed: $$ v = 1.30 \times 10^6 m/s $$ Thus, an electron would have to be moving at a speed of \(1.30 \times 10^6 m/s\) to have the same wavelength as the X-rays emitted by molybdenum.

Key concepts

Molecular Structure Determination

Understanding the complex world of molecules begins with deciphering their structures. This is where the science of molecular structure determination comes into play. Techniques such as X-ray diffraction are pivotal because they allow us to visualize something incredibly tiny and otherwise invisible to the naked eye.

X-rays are particularly suitable for this task due to their short wavelengths, which are of similar scale to the distances between atoms in a molecule. This compatibility allows X-rays to interact with the electrons cloud of the atoms and create a diffraction pattern that can be analyzed to reveal the positions of the atoms within the molecule.

To make this clearer, imagine trying to understand the details of a fingerprint; a light too broad won’t pick up the fine lines and ridges. Similarly, for studying molecules, we need a 'light' with a fine enough 'resolution', which is essentially what the short wavelength of X-rays provides.

Bragg's Law

In the world of crystallography, Bragg's Law is the star that outlines the relationship between the angles at which X-ray beams are diffracted and the spacing between the planes in the crystal lattice. Put simply, it's a powerful formula that helps us predict where the peaks and troughs of the diffraction pattern will occur, which is vital in determining the structure of crystals.

The equation given by Bragg’s Law is:
\[ n\beta = 2d\beta \text{sin}\theta \]
where \( n \) is the order of the reflected wave, \( \beta \) is the wavelength, \( d \) is the distance between the crystal planes, and \( \theta \) is the angle of incidence. It can be thought of as like hitting a 'sweet spot' angle where the reflections from different planes reinforce each other, resulting in a detectable diffraction. The precision that Bragg's Law provides is key in the intricate task of understanding the layout of atoms in a material.

de Broglie Wavelength

Delving into the quantum realm, we encounter the de Broglie wavelength. It's a distinctive concept that reveals the wave-like nature of particles. Essentially, de Broglie proposed that particles like electrons also display wave properties, and their wavelength is inversely proportional to their momentum.

The de Broglie wavelength is given by the equation:\[ \beta = \frac{h}{p} \]
where \( \beta \) is the wavelength, \( h \) is the Planck constant, and \( p \) is the momentum of the particle. This equation symbolizes the bridge between particle physics and wave mechanics, and it's crucial when you want to compare something we know behaves like a wave, such as X-rays, with particles like electrons.

Understanding de Broglie wavelength changes how we view particles, and it's fundamental in fields like quantum mechanics, allowing for predictions of behavior that seem almost magical from a classical perspective.