Why does an alpha particle traveling at the same speed as a beta particle have 'only' about twenty times the energy? Rather than over seven thousand? [closed]

Question

An alpha particle travels at about half the speed as a beta particle, right? ~5% of light speed versus ~10%?

Therefore, if you doubled its velocity, its energy would roughly quadruple, correct? (Or slightly more than quadruple due to a small Lorentz gamma factor at these slightly relativistic speeds...)

An average 0.1c beta particle has approximately 1 MeV, while the .05c alpha particle has about 5 MeV...

But an alpha particle has over 7000 times the mass, therefore if it traveled at the same velocity as a beta would possess over 7000 times the kinetic energy.... right? Not approximately 20?

What am I missing?

Answer 1

For some reason, I initially thought you were interested in kinetic energy, rather than total energy. I'll show the same analysis for both cases, since I think the plot for kinetic energy is more interesting.

Total energy

We can use the formula for relativistic energy

$$ E_t = \gamma mc^2 $$ where $\gamma=(1-v^2/c^2)^{-1/2}$ is the usual Lorentz factor and $m$ is the rest mass of the particle, to plot how energy depends on velocity for an alpha particle ($m_\alpha=3.73\ {\rm GeV}/c^2)$ and for a beta particle ($m_\beta=511\ {\rm keV}/c^2$)

In the part of the plot where both curves are approximately constant, the velocity is small enough that the particles are non-relativistic (since the energy is dominated by the rest energy). Then the total energies are approximately the rest masses times $c^2$, so the ratio of total energies is the ratio of rest masses.

In fact the fact that the ratio in energy at fixed velocity is the ratio in rest mass holds even for relativistic energies, which you can read off from the formula.

Kinetic energy

To look at kinetic energy, we can use the formula for relativistic kinetic energy, where one takes the total energy and subtracts the rest energy

$$ E_k = E_t - mc^2 = (\gamma-1) mc^2 $$ The resulting plot is

From here you should be able to read off the ratio in kinetic energy at any velocity you want. For example, at $v/c=0.2$, the ratio in kinetic energy is slightly less than $10^4$ (as a consistency check, the ratio in rest masses is $\sim 4\times 10^9 / 5\times 10^5$).

This is not surprising, since at fixed velocity, using the formula above, the ratio of kinetic energies is equal to the ratio in masses.

Answer 2

An alpha particle travels at about half the speed as a beta particle, right? ~5% of light speed versus ~10%?

If you have an accelerator, you can give your particles any speed you like. What comes out of a radioactive decay depends on the energy levels of the decaying nucleus and its daughters.

If I skim the Particle Data Group's table of commonly-used radioactive sources, the beta-decay energies are systematically lower than the alpha-decay energies. Most of the beta decays have maximum energies below 1 MeV, and therefore have mean beta energies below that (because the beta-decay energy spectrum is continuous). The alpha decays in that table, on the other hand, all seem to have energies in excess of 5 MeV.

If you are thinking of the classic "three species of radiation" discovery experiment, with the cathode-ray tube and the magnetic field which separates the radiation into its positive, negative, and neutral components, the most relevant source in that particular table is thorium-228. Thorium-228 undergoes a cascade of decays on its way to lead-208, emitting alpha particles with energies in the range 5–8 MeV, and beta particles with (mean) energies somewhere under 1 MeV.

A beta particle with kinetic energy $T=(\gamma-1)mc^2 \approx 1\rm\,MeV$ is unquestionably relativistic, with Lorentz factor $\gamma\approx 3$. From the definition $\gamma = \left(1-(v/c)^2\right)^{-1/2}$, we get a speed

$$ \begin{align} \left(\frac vc\right)^2 &= 1 - \frac 1{\gamma^2} \\ & \approx 1-\frac1{3^2} \approx \frac 8 9 \\ \left(\frac vc\right)^2 &\approx \big(0.94\big)^2 \end{align} $$

By contrast, an alpha particle with energy 8 MeV has the relativistic factor

\begin{align} \frac{(\gamma-1)mc^2}{mc^2} \approx \frac{8\rm\,MeV}{1000\rm\,MeV} &\approx 10^{-2} \\ \gamma &\approx 1 + 10^{-2} \end{align}

and therefore speeds like

\begin{align} \left(\frac vc\right)^2 &\approx 1 - \left(1+10^{-2}\right)^{-2} \\&\approx 1-\left(1 - 2\cdot 10^{-2}\right) \\ \frac vc &\approx \sqrt 2 \cdot 10^{-1} \approx 14\% \end{align}

Answer 3

Most beta particles travel much faster than $0.1c$. A typical beta particle of kinetic energy $1 \;\text{MeV}$ travels at $0.9c$. If it traveled at $0.1c$ (which is the slowest known speed in the case of $^{187}\text{Re}$), its kinetic energy will only be $2.5 \;\text{keV}$.