Question

Moessbauer spectroscopy is a technique for studying molecules by looking at a particular atom within them. For example, Moessbauer measurements of iron (Fe) inside hemoglobin, the molecule responsible for transporting oxygen in the blood, can be used to determine the hemoglobin's flexibility. The technique starts with X-rays emitted from the nuclei of \({ }^{57} \mathrm{Co}\) atoms. These X-rays are then used to study the Fe in the hemoglobin. The energy and momentum of each X-ray are \(14.0 \mathrm{keV}\) and \(14.0 \mathrm{keV} / \mathrm{c}\) (see Example 7.5 for an explanation of the units). \(A^{57}\) Co nucleus recoils as an X-ray is emitted. A single \({ }^{57}\) Co nucleus has a mass of \(9.52 \cdot 10^{-26}\) kg. What are the final momentum and kinetic energy of the \({ }^{57}\) Co nucleus? How do these compare to the values for the X-ray?

Short answer

**Short answer:** The final momentum of the ${ }^{57}\mathrm{Co}$ nucleus is equal and opposite to the initial momentum of the X-ray, with both having a magnitude of \(14.0 \mathrm{keV}/\mathrm{c}\). However, the final kinetic energy of the ${ }^{57}\mathrm{Co}$ nucleus (\(2.95 \times 10^{-6}\ \mathrm{keV}\)) is much smaller than the initial energy of the X-ray (\(14.0 \mathrm{keV}\)).

Step-by-step solution

Use the Conservation of Momentum formula to find the final momentum of the \({ }^{57}\mathrm{Co}\) nucleus

Since momentum is conserved in this process, the initial momentum is equal to the final momentum. The initial momentum of the X-ray is given as \(14.0 \mathrm{keV} / \mathrm{c}\). Since the \({ }^{57}\mathrm{Co}\) nucleus recoils, its final momentum will be equal and in the opposite direction of the X-ray's initial momentum. Thus, the final momentum of the \({ }^{57}\mathrm{Co}\) nucleus is \(-14.0 \mathrm{keV}/\mathrm{c}\).

Convert the momentum of the \({ }^{57}\mathrm{Co}\) nucleus to SI units

Since we will be working with the momentum in terms of mass and velocity, we need to convert the momentum from keV/c to kg m/s. To do this, we first convert keV to joule (J) using the following conversion formula: 1 keV = \(1.602 \times 10^{-16}\) J So, \(-14.0 \mathrm{keV} = -14.0 \times 1.602 \times 10^{-16}\ \mathrm{J}\) Then, since \(p = \frac{E}{c}\), multiply the energy in J by c to get momentum in kg m/s: \(p_{57Co} = -14.0 × 1.602 × 10^{-16}\ \mathrm{J} × c = -2.996 \times 10^{-23}\ \mathrm{kg}\ \mathrm{m/s}\)

Calculate the velocity of the \({ }^{57}\mathrm{Co}\) nucleus

Now, we have the momentum of the \({ }^{57}\mathrm{Co}\) nucleus in kg m/s. We can use the formula \(p = m \times v\) to find the velocity of the \({ }^{57}\mathrm{Co}\) nucleus: \(v = \frac{p}{m} = \frac{-2.996 \times 10^{-23}\ \mathrm{kg}\ \mathrm{m/s}}{9.52 \times 10^{-26}\ \mathrm{kg}} = -3.15 \times 10^2\ \mathrm{m/s}\)

Calculate the kinetic energy of the \({ }^{57}\mathrm{Co}\) nucleus

We will use the kinetic energy formula, \(KE = \frac{1}{2}mv^2\), to calculate the kinetic energy of the \({ }^{57}\mathrm{Co}\) nucleus: \(KE_{57Co} = \frac{1}{2} (9.52 \times 10^{-26}\ \mathrm{kg}) (-3.15 \times 10^2\ \mathrm{m/s})^2 = 4.72 \times 10^{-22}\ \mathrm{J}\) To compare the kinetic energy to the energy of the X-ray, convert the energy back to keV: \(KE_{57Co} = 4.72 \times 10^{-22}\ \mathrm{J} \times \frac{1}{1.602 \times 10^{-16}\ \mathrm{J/keV}} = 2.95 \times 10^{-6}\ \mathrm{keV}\)

Compare the values of momentum and kinetic energy

Now that we have the final momentum and kinetic energy of the \({ }^{57}\mathrm{Co}\) nucleus, let's compare these values to the X-ray's initial energy and momentum: X-ray momentum: \(14.0 \mathrm{keV} / \mathrm{c}\) \({ }^{57}\mathrm{Co}\) nucleus final momentum: \(-14.0 \mathrm{keV} / \mathrm{c}\) (equal and opposite) X-ray energy: \(14.0 \mathrm{keV}\) \({ }^{57}\mathrm{Co}\) nucleus final kinetic energy: \(2.95 \times 10^{-6}\ \mathrm{keV}\) (much smaller)

Key concepts

Momentum Conservation

Understanding momentum conservation is crucial in physics, especially when analyzing interactions between objects. In our Moessbauer spectroscopy exercise involving the emission of an X-ray from a ⁵⁷Co nucleus, momentum conservation plays a central role.

When the X-ray is emitted, the nucleus of the cobalt atom recoils to maintain the system's overall momentum. Since no external forces are acting on the nucleus-X-ray system, the total momentum before and after the emission must be equal. The X-ray carries away momentum, and in response, the cobalt nucleus moves in the opposite direction with an equal amount of momentum, ensuring the conservation of momentum within the isolated system.

This principle is the reason why the final momentum of the ⁵⁷Co nucleus is exactly opposite to the momentum of the emitted X-ray.

Kinetic Energy Calculation

After establishing the recoil momentum of the ⁵⁷Co nucleus, we can calculate its kinetic energy. Kinetic energy, the energy possessed by an object due to its motion, is given by the formula
KE = \( \frac{1}{2}mv^2 \).

To find the kinetic energy of the ⁵⁷Co nucleus, we utilize its mass and the velocity we calculated from its momentum. Importantly, the kinetic energy depends on the square of the velocity, so even though the nucleus moves in the opposite direction, the negative sign in velocity has no effect on kinetic energy.

As seen in the solution steps, the kinetic energy of the ⁵⁷Co nucleus is markedly smaller than the energy of the X-ray. This discrepancy highlights an interesting aspect of interactions on the atomic scale: small masses like individual nuclei possess significantly less kinetic energy for the same momentum compared to much less massive particles like photons.

Nuclear Recoil

Nuclear recoil is the term used to describe the backward movement of a nucleus after it emits a particle, such as an X-ray photon in Moessbauer spectroscopy. This phenomenon results from the conservation of momentum where the nucleus and the X-ray photon together constituted a closed system.

The recoil energy is often quite small due to the relative mass difference between the nucleus and the emitted photon; however, it is still measurable and can impact the energy of emitted or absorbed photons when the nuclide is part of a solid structure, as with the iron in hemoglobin. This impact is key to understanding how the Moessbauer spectroscopy technique can be used to measure the physical properties of molecules, and how recoiling nuclei behave in different environments.

Physics of Hemoglobin

In Moessbauer spectroscopy, the physics of hemoglobin becomes relevant when we explore how the iron nuclei within hemoglobin molecules interact with X-rays. Hemoglobin, a complex protein in our red blood cells, uses iron atoms to bind oxygen and transport it throughout the body.

The flexibility and environments of these iron atoms affect their nuclear recoil when exposed to gamma radiation, like the X-rays from a ⁵⁷Co source used in the Moessbauer effect. By analyzing the resulting spectra, scientists can infer the properties of hemoglobin, such as its ability to bind with oxygen, which is influenced by the structure and dynamics of the iron-containing heme group.

This intricate relationship between physics at an atomic level and biological functionality is just one example of how Moessbauer spectroscopy provides insights into molecular biology. It's a powerful demonstration of the unity of physics and chemistry in understanding life processes.