Differences
This shows you the differences between two versions of the page.
Next revision | Previous revision | ||
183_notes:rest_mass [2014/10/03 20:28] – created caballero | 183_notes:rest_mass [2021/05/06 20:02] (current) – [Example: Neutron Decay] stumptyl | ||
---|---|---|---|
Line 1: | Line 1: | ||
===== Change of Rest Mass Energy ===== | ===== Change of Rest Mass Energy ===== | ||
- | Until now, you have dealt with particles that do not change their identity. Changing the identity of a particle occurs when a [[http:// | + | Until now, you have dealt with particles that do not change their identity. Changing the identity of a particle occurs when a [[http:// |
+ | ** | ||
==== The Electron Volt ==== | ==== The Electron Volt ==== | ||
For many situations, the unit of the Joule is quite useful. For very small particles like neutrons, protons, and electrons, a different unit is used typically. Consider the rest mass energy for the neutron, | For many situations, the unit of the Joule is quite useful. For very small particles like neutrons, protons, and electrons, a different unit is used typically. Consider the rest mass energy for the neutron, | ||
- | $$E_{rest} = mc^2 = (1.6749\times10^{-27}kg)(3\times10^{8})^2 = 1.51\times10^{-10}J$$ | + | $$E_{rest} = mc^2 = (1.6749\times10^{-27}kg)(3\times10^{8} |
+ | |||
+ | This energy is quite small, so folks often convert this to the electron volt which is, | ||
+ | |||
+ | $$1eV = 1.6\times10^{-19}J$$ | ||
+ | |||
+ | The electron volt is just another unit of energy. It can be used to scale the rest mass energy of the neutron. | ||
+ | |||
+ | $$E_{rest} = 1.51\times10^{-10}J \dfrac{1eV}{1.6\times10^{-19}J} = 9.396 \times 10^8 eV = 939.6 MeV$$ | ||
+ | |||
+ | Typically, elementary particle rest mass energies are given in " | ||
+ | |||
+ | ^ Particle | ||
+ | | Neutrino, $\nu$ | $\approx$ 0 MeV | | ||
+ | | Electron, $e^-$ | 0.511 MeV | | ||
+ | | Proton, | ||
+ | | Neutron, | ||
+ | |||
+ | |||
+ | ==== Example: Neutron Decay ==== | ||
+ | |||
+ | [{{ 183_notes: | ||
+ | As an example of the change of particle identity, consider [[http:// | ||
+ | |||
+ | The system before the decay consists of just the neutron. After the decay, let's choose the system to be the proton, electron, and anti-neutrino. If that's what we choose for the system, there' | ||
+ | |||
+ | - System: neutron (before decay); proton, electron, and anti-neutrino (after decay) | ||
+ | - Surroundings: | ||
+ | |||
+ | So you can apply the [[183_notes: | ||
+ | |||
+ | $$E_{sys,f} = E_{sys,i} + W$$ | ||
- | ==== Neutron Decay ==== | + | The system energies consist of the sum of the rest mass energies and the kinetic energies of the particles. |
+ | $$(m_pc^2 +K_p) + (m_ec^2 + K_e) + K_{\bar{\nu}} = (m_nc^2 + K_n) + W $$ | ||
+ | $$(m_pc^2 +K_p) + (m_ec^2 + K_e) + K_{\bar{\nu}} = (m_nc^2 + 0) + 0 $$ | ||
+ | $$(m_pc^2 + m_ec^2) +K_p + K_e + K_{\bar{\nu}} = m_nc^2 $$ | ||
+ | $$(m_pc^2 + m_ec^2) + (K_p + K_e + K_{\bar{\nu}}) = m_nc^2 $$ | ||
+ | $$K_p + K_e + K_{\bar{\nu}} = m_nc^2 | ||
+ | $$K_p + K_e + K_{\bar{\nu}} = 939.6 MeV - (938.3 MeV + 0.511 MeV)$$ | ||
+ | $$K_p + K_e + K_{\bar{\nu}} = 0.8MeV$$ | ||
+ | This energy is available to the products for their motion. This decay must also [[183_notes: |