Speed and Momentum of a Beta Particle

AI Thread Summary
The discussion revolves around calculating the speed and momentum of a beta particle emitted from a 14C nucleus. The initial calculations were based on non-relativistic equations, yielding an incorrect speed of 2.31 x 10^8 m/s, which exceeds the speed of light. Participants emphasize the necessity of using relativistic formulas for energy due to the high kinetic energy of 127 keV. The correct approach involves applying relativistic equations to accurately determine the speed and momentum of both the beta particle and the remaining nucleus. The conversation highlights the importance of recognizing when to switch from classical to relativistic physics in particle physics problems.
ihearyourecho
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Homework Statement



A 14C nucleus, initially at rest, emits a beta particle. The beta particle is an electron with 127 keV of kinetic energy.

A) What is the speed of the beta particle?
B) What is the momentum of the beta particle?
C) What is the momentum of the nucleus after it emits the beta particle?
D) What is the speed of the nucleus after it emits the beta particle?

Homework Equations



v = √[ 2K / m]

The Attempt at a Solution



Okay, so this is my work, but it gives the wrong answer.

Kinetic energy of beta particle K = 127 keV = 127 * 10 ^ 3 eV
= 127 * 10 ^ 3 * 1.6 * 10 ^ -19 J
= 203.2* 10^ -16 J

K = ( 1/ 2) mv ^ 2
v = √[ 2K / m]
m=9.11* 10 ^ -31 kg
v = 2.31* 10 ^ 8 m / s
THIS IS NOT RIGHT.

Can someone help me out where my thinking was flawed please?
 
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Hi ihearyourecho! :smile:

(try using the X2 tag just above the Reply box :wink:)
ihearyourecho said:
K = ( 1/ 2) mv ^ 2
v = √[ 2K / m]
m=9.11* 10 ^ -31 kg
v = 2.31* 10 ^ 8 m / s
THIS IS NOT RIGHT.

The speed of light is c = 3*108 m/s,

so your v is about 0.77 c, which is far too large to be able to use non-relativistic formulas. :redface:

You need the relativistic formula for energy. :wink:​
 

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