Relationship of velocity of an electron/velocity of proton to mass ratio

Click For Summary
The discussion focuses on deriving the ratio of the velocities of an electron and a proton after being accelerated through the same electric potential difference. Using the equation for kinetic energy, participants clarify that the velocity of each particle can be expressed as a function of its charge and mass. The correct expressions for the speeds are v_e^2 = -2q_eΔV/m_e for the electron and v_p^2 = -2q_pΔV/m_p for the proton. By taking the ratio of these two equations, the relationship between the velocities and masses can be established. The conversation emphasizes the importance of correctly applying the conservation of energy principle to arrive at the desired ratio.
sheri1987
Messages
47
Reaction score
0
Relationship of velocity of an electron/velocity of proton to...mass ratio

Homework Statement



An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed ve, while the proton acquires a speed vp.


--------------------------------------------------------------------------------

(d) What is the algebraic expression for the ratio ve/vp of the speed of the electron to that of the proton? Express your answer in terms of the mass me of the electron and the mass mp of the proton. (Answer using m_e for me and m_p for mp.)

Homework Equations



SO I need to write an equation relating Velocity of electron/ velocity of proton...I understand that much...but I am unsure where to start.

I think I need the equation 1/2mv^2 = -q (charge)deltaV(voltage)


The Attempt at a Solution



by using the equation above I solved and got sqrt(2qdeltaV/m) ...I am not sure where to go next...? Can anyone help me out?
 
Physics news on Phys.org
sheri1987 said:
I think I need the equation 1/2mv^2 = -q (charge)deltaV(voltage)

Yes, that's what you want to use.

by using the equation above I solved and got sqrt(2qdeltaV/m) ...I am not sure where to go next...? Can anyone help me out?

Actually, you should get sqrt(-2qdeltaV/m). Then if q<0 (like it is for an electron) then deltaV>0 and the radicand is positive. And if q>0 (like it is for the proton) then deltaV<0, and again the radicand is positive. Just find expressions for the speeds of the proton and the electron, and take their ratio and you'll have it.
 
sheri1987 said:
by using the equation above I solved and got sqrt(2qdeltaV/m) ...I am not sure where to go next...? Can anyone help me out?
So you have used conservation of energy to almost correctly [note the sign correction] solved for the velocity of a generic particle of charge q and mass m (traveling at non-relativistic speeds). Let me write your equation out in a form that may be easier to work with;

v^2 = -2q\Delta V \cdot \frac{1}{m}

So now you want that ratio v_e/v_p. So you can now write two equations;

v_e^2 = -2q_e\Delta V \cdot \frac{1}{m_e}

v_p^2 = -2q_p\Delta V \cdot \frac{1}{m_p}

Can you take the next step?

Edit: Tom beat me to it!
 
Last edited:
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K