 # Relativistic mass = m_0 gamma where gamma is the lorentz factor. So if i have a mass that is .5 at rest then it is safe to say that the relativistic mass will be 1 if it goes at sqrt(3)/2 c. Wwhat happens if that .5 is actually a radioactive isotope and is decaying while speeding up? Then at what speed will it approximately 'equal' 1? Brandon Monroe 2022-08-16 Answered
Relativistic mass = ${m}_{0}\gamma$ where $\gamma$ is the lorentz factor.
So, if a mass that is $.5$ at rest then it is safe to say that the relativistic mass will be $1$ if it goes at $\frac{\sqrt{3}}{2}c$.
What happens if that $.5$ is actually a radioactive isotope and is decaying while speeding up? Then at what speed will it approximately equal $1$?
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Radioactive decay doesn't appreciably change the mass of most isotopes. So for practical purposes, you can ignore the radioactivity when calculating relativistic mass increases.
If you did want to take into account the very slight change in mass from radioactive decay, assuming you treat any emitted alpha or beta particles as lost mass, you'd just replace the constant ${m}_{0}$ with the variable mass of the radioactive isotope as it decays.

We have step-by-step solutions for your answer! Filipinacws
${m}_{e0}\simeq 0.511\frac{Mev}{{c}^{2}}$
this case could be electron mass. when relativistic factor is equal two $\gamma =2$ which means the speed parameter of electron is equal to $\beta =\frac{\sqrt{3}}{2}$. so relativistic electron will be:
${m}_{e}={m}_{e0}.\gamma \simeq 1.0\frac{Mev}{{c}^{2}}$
that means kinetic energy of electron is equal to rest energy of electron
${k}_{e}={m}_{e}-{m}_{e0}={m}_{e0}\left(2-1\right)={m}_{e0}$
and momentum of electron will be:
${p}_{e}={m}_{e0}\sqrt{3}.c\simeq 0.885\frac{Mev}{c}$

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