Question

An excited state \(X^{*}\) of an atom at rest drops to its ground state \(X\) by emitting a photon. In atomic physics it is usual to assume that the energy \(E_{\gamma}\) of the photon is equal to the difference in energies of the two atomic states, \(\Delta E=\left(M^{*}-M\right) c^{2},\) where \(M\) and \(M^{*}\) are the rest masses of the ground and excited states of the atom. This cannot be exactly true, since the recoiling atom X must carry away some of the energy \(\Delta E .\) Show that in fact \(E_{\gamma}=\Delta E\left[1-\Delta E /\left(2 M^{*} c^{2}\right)\right] .\) Given that \(\Delta E\) is of order a few ev, while the lightest atom has \(M\) of order \(1 \mathrm{GeV} / c^{2},\) discuss the validity of the assumption that \(E_{\gamma}=\Delta E\)

Short answer

The assumption \(E_{\gamma} = \Delta E\) is valid because the correction term is negligible due to \(\Delta E\) being much smaller than \(M^{*}c^2\).

Step-by-step solution

Understand Energy Conservation

First, note that energy conservation requires the total energy before and after emission to remain constant. Before emission, the energy is the rest energy of the excited atom, which is \( M^{*} c^{2} \). After emission, the energy comprises the photon energy \( E_{\gamma} \) and the kinetic energy of atom X. Thus, the total energy conservation equation is:\[ M^{*} c^{2} = E_{\gamma} + K \] where \( K \) is the kinetic energy of the recoiling atom.

Recognize Momentum Conservation

Recall that momentum must also be conserved. Before emission, the atom has zero momentum at rest. After emission, the atom and photon must have equal and opposite momenta. If \( p_{\gamma} \) is the photon momentum, then the atom's momentum \( p_{X} \) must be equal in magnitude:\[ p_{X} = p_{\gamma} = \frac{E_{\gamma}}{c} \] based on the relationship \( E_{\gamma} = p_{\gamma} c \) for photons.

Express Recoiling Atom's Kinetic Energy

The kinetic energy \( K \) of the recoiling atom can be expressed in terms of its momentum:\[ K = \frac{p_{X}^{2}}{2M} = \frac{E_{\gamma}^{2}}{2Mc^2} \] This relationship allows us to substitute back into the energy conservation equation.

Substitute and Simplify

Replace \( K \) in the energy conservation equation:\[ M^{*} c^{2} = E_{\gamma} + \frac{E_{\gamma}^{2}}{2M^{*}c^2} \] Rearrange terms to find \( E_{\gamma} \) in terms of \( \Delta E = M^{*} c^{2} - M c^{2} \):\[ E_{\gamma} \left( 1 + \frac{E_{\gamma}}{2M^{*}c^2} \right) = \Delta E \]Assuming \( E_{\gamma} \approx \Delta E \), approximate further to derive:\[ E_{\gamma} = \Delta E \left( 1 - \frac{\Delta E}{2M^{*}c^2} \right) \] reflecting the energy carried away by the recoiling atom.

Discuss Validity of Assumption

Since \( \Delta E \) is much smaller than \( M^{*} c^{2} \) (\( eV \) compared to \( GeV \)), the correction term \( \frac{\Delta E}{2M^{*}c^2} \) is negligible. Thus, the assumption \( E_{\gamma} \approx \Delta E \) is valid for most practical purposes, as the difference is minimal and unlikely to impact most experimental results.

Key concepts

Energy Conservation

Energy conservation is a fundamental principle in physics stating that the total energy of an isolated system remains constant. In the context of atomic physics, when an atom transitions from an excited state to its ground state by emitting a photon, energy conservation dictates that the total energy before and after the emission process must be equal.

In the given exercise, before the photon emission, the total energy is the rest energy of the excited atom, denoted by \( M^{*} c^{2} \). After the emission, the energy consists of two parts: the energy of the emitted photon \( E_{\gamma} \) and the kinetic energy \( K \) of the recoiling atom. Hence, the equation of energy conservation in this scenario can be written as:

• \( M^{*} c^{2} = E_{\gamma} + K \)

The exercise highlights that not all the energy from the excited state goes into the photon; some energy is used for the recoil of the atom. This energy distribution ensures the total energy remains unchanged, consistent with the law of energy conservation.

Moreover, the difference between the two states' energies is represented as \( \Delta E = (M^{*} - M)c^{2} \), where both potential and kinetic energies are taken into account in the total energy budget.

Momentum Conservation

Momentum conservation is equally crucial in understanding photon emission processes in atomic physics. It states that the total momentum of a closed system remains constant over time. In our problem, the atom initially at rest has zero momentum before photon emission. However, after the emission, both the photon and the atom have momenta that satisfy the momentum conservation law.

The emitted photon carries away a momentum \( p_{\gamma} \), which, due to conservation, must be equal and opposite to the momentum of the recoiling atom \( p_{X} \). This relationship is grounded in the principle that in absence of external forces, changes in momentum within the system must balance out:

• \( p_{X} = p_{\gamma} = \frac{E_{\gamma}}{c} \)

For a photon, its energy and momentum are connected by the equation \( E_{\gamma} = p_{\gamma} c \). This makes it possible to substitute the momentum-related terms into the equations of motion to analyze kinetic energy and photon energy relations.

Understanding momentum conservation alongside energy conservation in photon emission processes not only reinforces the physics behind atomic transitions but also prepares one to analyze more complex systems involving particle interactions. It shows how energy and momentum interplay to uphold conservation laws in physics.

Photon Emission

Photon emission is a key process in atomic physics where an excited atom releases energy in the form of a photon when it returns to a lower energy state. The calculation in the given exercise illustrates how the energy of the emitted photon \( E_{\gamma} \) slightly deviates from the simple energy difference \( \Delta E \) due to the recoil effect of the atom.

This deviation is captured in the relationship:

• \( E_{\gamma} = \Delta E \left( 1 - \frac{\Delta E}{2M^{*}c^2} \right) \)

The term \( \Delta E \) represents the theoretical energy difference between the excited and ground states, while the factor \( \left( 1 - \frac{\Delta E}{2M^{*}c^2} \right) \) accounts for the minor energy lost due to the atom's recoil.

Since \( \Delta E \) (typically measured in electronvolts) is much smaller than \( M^{*} c^{2} \) (typically measured in giga-electronvolts for atoms), the correction term \( \frac{\Delta E}{2M^{*}c^2} \) becomes negligible. Thus, in practical terms, the energy of the photon is very close to \( \Delta E \), affirming the approximation used in many experimental analyses.

By understanding the photon emission process, students learn how fundamental processes in atomic physics align with the principles of conservation, and why approximations are often justified based on the scale of values involved.