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 \({ }^{57}\) Co atoms. These X-rays are then used to study the Fe in the hemoglobin. The energy and momentum of each X-ray are \(14 \mathrm{keV}\) and \(14 \mathrm{keV} / \mathrm{c}\) (see Example 7.5 for an explanation of the units). \(\mathrm{A}^{57}\) Co nucleus recoils as an \(\mathrm{X}\) -ray is emitted. A single \({ }^{57}\) Co nucleus has a mass of \(9.52 \cdot 10^{-26} \mathrm{~kg} .\) What are the final momentum and kinetic energy of the \({ }^{57}\) Co nucleus? How do these compare to the values for the X-ray?

Step by step solution

01

Identify the given values in the problem

We know the energy and momentum of each X-ray are \(14 \mathrm{keV}\) and \(14 \mathrm{keV} / \mathrm{c}\), and the mass of a single \({ }^{57}\) Co nucleus is \(9.52 \cdot 10^{-26} \mathrm{~kg}\).

02

Find the momentum of the \({ }^{57}\) Co nucleus after the X-ray emission

By conservation of momentum, the final momentum of the nucleus must be equal in magnitude and opposite in direction to the momentum of the emitted X-ray. Therefore, the momentum of the \({ }^{57}\) Co nucleus after the emission is \(-14 \mathrm{keV} / \mathrm{c}\) (note the negative sign).

03

Convert the units to SI for further calculations

Convert the energy of the emitted X-ray from keV to joules using the conversion factor: 1 keV = 1.602 × 10^{-16} J. Then, we have: Energy of X-ray: \(14 \mathrm{keV} \times (1.602 \times 10^{-16} \mathrm{J} / \mathrm{keV}) = 2.243 \times 10^{-15} \mathrm{J}\) Now, we can find the momentum of the emitted X-ray in SI units as well: P_momentum = \((14 \mathrm{keV}/ \mathrm{c}) \times 1.602 \times 10^{-16} / \mathrm{keV} = (2.243 \times 10^{-15}\mathrm{~J}) / 2.998 × 10^{8} \mathrm{m/s}\) We also know that the mass of the \({ }^{57}\) Co nucleus is \(9.52 \cdot 10^{-26} \mathrm{~kg}\).

04

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

Now that we have the momentum and mass of the \({ }^{57}\) Co nucleus, we can find the recoil velocity using the momentum formula (p = mv): \(v_\text{recoil} = \frac{p_\text{momentum}}{m_\text{nucleus}} = \frac{-(2.243 \times 10^{-15}\mathrm{~J}) / 2.998 × 10^{8} \mathrm{m/s}}{9.52 × 10^{-26}\mathrm{~kg}}\)

05

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

We can now use the relationship between kinetic energy, mass, and velocity (KE = 1/2 mv²) to determine the kinetic energy of the \({ }^{57}\) Co nucleus: \(KE_\text{nucleus} = \frac{1}{2} m_\text{nucleus} v_\text{recoil}^2\) Plug in the mass of the nucleus and the recoil velocity calculated in Step 4 to find the kinetic energy.

06

Compare the kinetic energy and momentum of the nucleus with the X-ray

Finally, compare the kinetic energy and momentum of the \({ }^{57}\) Co nucleus after the emission to the initial energy and momentum of the emitted X-ray: 1. Kinetic energy: We calculated the kinetic energy of the nucleus in Step 5 and found that of the X-ray in Step 3. Compare these values. 2. Momentum: The final momentum of the nucleus is \(-14 \mathrm{keV} / \mathrm{c}\) (found in Step 2), while the initial momentum of the X-ray is \(14 \mathrm{keV} / \mathrm{c}\). Compare these values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nuclear Recoil

When an atom emits an X-ray, it experiences what's known as nuclear recoil. Think of it like the kickback you feel when you push a skateboard away from you — the skateboard moves in one direction, and your body moves in the opposite direction. This is a result of the conservation of momentum, which holds that the total momentum of a system remains constant if there are no external forces.

In Moessbauer spectroscopy, a uclear recoil} occurs in the ucleus} of an atom after an X-ray is emitted. If an X-ray is emitted from a ucleus, the nucleus itself must move in the opposite direction to maintain the system's overall momentum. The recoil momentum of a uclear is usually quite small because the nucleus is much more massive than the emitted X-ray photon.

Conservation of Momentum

Conservation of momentum is a fundamental principle of physics stating that the total momentum of a closed system is constant if no external forces are acting on it. In our case, it explains the behavior of the ucleus when an X-ray is emitted.

For Moessbauer spectroscopy, when the ucleus recoils, it is a practical display of the conservation of momentum. The emitted X-ray and recoiling nucleus essentially exchange momentum so that the sum of their momenta before and after the emission is the same. This gives us a way to track the movement and properties of particles within atoms, offering deep insights into their behavior.

Kinetic Energy Calculations

Kinetic energy is the energy a body possesses by virtue of being in motion. After figuring out the recoil velocity of a ucleus from conservation of momentum equations, as shown in the steps of the solution, we need to calculate its kinetic energy.

The formula for kinetic energy, \( KE = \frac{1}{2} mv^2 \), relates the mass (\( m \)) of an object and the square of its velocity (\( v \)). In the case of the recoiling ucleus, the kinetic energy is incredibly small compared to the emitted X-ray because the nucleus's velocity is tiny. This low kinetic energy also explains why Moessbauer spectroscopy can be so precise — the ucleus stays more or less in the same place despite the emission.

X-ray Emission

X-ray emission is one of the primary events in the analysis performed by Moessbauer spectroscopy. X-rays are electromagnetic waves with very high energy, and when they're emitted from a ucleus, they provide valuable information about the structure and properties of the material being studied.

The energy of the emitted X-ray can be precisely measured, which in turn allows scientists to make exact inferences about the emitting ucleus's environment. The Moessbauer effect is unique in that it gives us a glimpse into the tiny shifts in energy resulting from different bonding environments within molecules, such as the hemoglobin mentioned in the exercise.