Determine type of particle emitted from decay?

[bob dobilina]

Homework Statement

A Phosphorous 34 decays and emits a particle. A JJ Thomson experiment is done to find out the charge to mass ratio of this particle. The particle moves undeflected through mutually perpendicular magnetic and electric fields of 2.00 x 10^-3 T and 1.08 x 10⁴ N/C, respectively. When the electric field is turned off, the particle deflects to a radius of 1.53x10^-2m. Determine the type of particle emitted.

Homework Equations

Fe = electric Force
Fm = Magnetic Force
Fc= centripetal Force
B=Magnetic Field
r=radius of curvature
m=mass
v=velocity
E=Electric Field
q=charge of particle

Fe= qE
Fm=qvB
Fc=mv²/r

When the electric field is turned off we know that Fm=Fc
Because the particle is undeflected, we know that Fe=Fm

The Attempt at a Solution

To find the v of this particle, we can manipulate the formual of Fe=Fm into:
E=vB
v=(E/B)

To find the mass of the particle we can manipulate the formual of Fm=Fc into:
m= (Fc x r)/v²

So, i figured out the velocity and mass of the particle, and now I am trying to determine the type of particle emitted. Am i able to do this using mass only? Or should I try and find the charge of the particle, and if so how?

Thanks

 

[BvU]

You have no info to help you find the charge. So you are restricted to the type of particle (pion, muon, Kaon, electron/positron), just like the exercise text puts it.

 

[bob dobilina]

You have no info to help you find the charge. So you are restricted to the type of particle (pion, muon, Kaon, electron/positron), just like the exercise text puts it.

Would i be able to manipulate Fm=Fc to find q (the charge).
qvB=mv²/r
q=(mv)/(rB)
?

 

[BvU]

If you have the direction of $\vec B$, $\vec v$ and know which way it deflects, yes. But you don't.

 

[bob dobilina]

If you have the direction of $\vec B$, $\vec v$ and know which way it deflects, yes. But you don't.

Alright. I figure the charge to mass ratio should be:
Fm=Fc
qvB = (mv²)/r
(q/m)=v/(Br)
When i crunch the numbers I get an answer of 1.7647 x 10 ¹¹
Same as an electron..what do you think of this?

 

[BvU]

Crunching numbers gives a number. But you need a mass. In kilograms, preferably (not in stones, lbs or that kind of stuff).

Pretty heavy electrons ! perhaps $10^{-11}$ (if the 'number' is in kilograms) ?



But I think you are doing fine. Click 34P in this table to see the decay mode...[edit] Oops I forgot, $m_e = 9.10938291 × 10^{-31}$ kilograms ??

[edit] Oops² I remember the value of e/m in C/kg is the same as your number, that's a lot better !

 

[bob dobilina]

Crunching numbers gives a number. But you need a mass. In kilograms, preferably (not in stones, lbs or that kind of stuff).

Pretty heavy electrons ! perhaps $10^{-11}$ (if the 'number' is in kilograms) ?



But I think you are doing fine. Click 34P in this table to see the decay mode...[edit] Oops I forgot, $m_e = 9.10938291 × 10^{-31}$ kilograms ??

[edit] Oops² I remember the value of e/m in C/kg is the same as your number, that's a lot better !

Sorry I should have included the units in my ratio. Awesome. Thank you for the help.