Voltage that gives acceleration to relativistic particles

[Antoha1]

Homework Statement

A proton and an alpha particle begin to move in an electric field with a potential difference (voltage) of U.
After passing through this field, the mass of the proton is three times less than the mass of the alpha particle. What is this potential difference?

Relevant Equations

q is charge;

Applying conservation of energy law:
Both particles got energy $A$ from eletric fiel: $A_{p}=eU$ and $A_{\alpha}=2eU$.
Lets say particles weren't moving before accelerating them.

1)
Proton energy: $E_{p}=m_{p}c^2+eU$
Alpha particle energy: $E_{\alpha}=m_{\alpha}c^2+2eU$

Before acceleration $\frac{m_{p}}{m_{\alpha}} \approx \frac{1}{4}$

2) After acceleration:
$E_{p}=\gamma_{p}m_{p}c^2$
$E_{\alpha}=\gamma_{\alpha}m_{\alpha}c^2$

But it is said that after acceleration $\frac{M_{\alpha}}{M_{p}}=3$

So, how can I solve for U, I don't know how to deal with this rest mass relation and relativistic mass relation also thinking if I extract gamma, speeds of the particles won't cancel out and I'd be left with two more not known values. Maybe different way of solving there is?

 

[jbriggs444]

Homework Statement: A proton and an alpha particle begin to move in an electric field with a potential difference (voltage) of U.
After passing through this field, the mass of the proton is three times less than the mass of the alpha particle. What is this potential difference?

When the question asks for mass, do you understand it to mean relativistic mass or invariant mass?

Obviously one can discard the idea that invariant mass is meant immediately. The invariant masses are always in the ratio of one to four regardless of velocity. Never one to three.

You understand that relativistic mass is just another name for total energy, right? Almost certainly so since you've provided the relevant formulas.

Relevant Equations: q is charge;

Applying conservation of energy law:
Both particles got energy $A$ from eletric fiel: $A_{p}=eU$ and $A_{\alpha}=2eU$.
Lets say particles weren't moving before accelerating them.

1)
Proton energy: $E_{p}=m_{p}c^2+eU$
Alpha particle energy: $E_{\alpha}=m_{\alpha}c^2+2eU$

Before acceleration $\frac{m_{p}}{m_{\alpha}} \approx \frac{1}{4}$

You have all of the pieces you need right there. You have formulas for the final energies. Two formulas with one unknown: $U$. You have been told that the final energies are in the ratio of one to three. You can look up the invariant mass of a proton (or of an alpha) in electron volts.

You should be able to solve for U (in Volts) without worrying about finding a value for either gamma.

 

[Antoha1]

do I understand correctly: Relativistic mass is just total energy so the energy ratio is 3Ep=Ea and I can manage everything with 2 first equations

 

[jbriggs444]

do I understand correctly: Relativistic mass is just total energy so the energy ratio is 3Ep=Ea and I can manage everything with 2 first equations

Yes.