A ray photon emitted by ^22688 Ra has an energy of 0.186...

A ray photon emitted by $^{226}_{88}$ Ra has an energy of 0.186 MeV. Use conservation of linear momentum to calculate the recoil speed of a $_{88}^{226} \mathrm{Ra}$ nucleus after such a $\gamma$ ray is emitted. Assume that the nucleus is at rest initially, and that relativistic effects can be ignored.

Key Concepts

Conservation of Linear Momentum

This fundamental principle states that within an isolated system where no external forces are acting, the total momentum remains constant. In the context of photon emission, it implies that the momentum carried away by the photon must be balanced by the recoil momentum of the emitting nucleus, ensuring the net momentum remains zero when the system was initially at rest.

Photon Energy-Momentum Relationship

Photons, even though they have no rest mass, possess momentum that is directly related to their energy by the relation p = E/c, where E is the energy of the photon and c is the speed of light. This relationship is essential when calculating the momentum transferred to the nucleus during photon emission.

Non-Relativistic Approximation

When the speeds involved in a problem are much smaller than the speed of light, relativistic effects can be neglected, allowing the use of classical mechanics. This approximation justifies using the equation v = p/m for the recoil speed, where p is the momentum and m is the mass of the nucleus.

Nuclear Recoil

Nuclear recoil describes the backward motion of a nucleus as it emits a particle or radiation, such as a gamma ray. This recoil is a direct consequence of momentum conservation, where the emitted photon's momentum results in a small but calculable velocity imparted to the nucleus.

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