How Do You Calculate the Potential Difference to Accelerate a He+ Ion?

[BuBbLeS01]

Electric Potential Difference...Please HELP!

Homework Statement

What potential difference is needed to accelerate a He+ ion (charge +e, mass 4u) from rest to a speed of 2690000 m/s?

Homework Equations

The Attempt at a Solution

Change in V = Change in K
Change in V = 1/2 mv^2
But what dose mass = 4u mean? What is u?

 

[rohanprabhu]

But what dose mass = 4u mean? What is u?

http://en.wikipedia.org/wiki/Atomic_mass_unit

 

[BuBbLeS01]

Oh ok...
so is the rest of the problem set up correctly?
So I do...
1/2 * (4 * 1.66x10^-27) * (2690000^2) = 2.4x10^-14 V

 

[rohanprabhu]

Oh ok...
so is the rest of the problem set up correctly?
So I do...
1/2 * (4 * 1.66x10^-27) * (2690000^2) = 2.4x10^-14 V

actually.. i think you have something fundamentally wrong with:

change in V = change in K

this is because, the equation is dimensionally incorrect. 'K' or it's change, has the units of energy, whereas 'V' has the units of 'Energy/Charge'.

your solution maybe right.. or i don't know what.. just explain to me why you took the above mentioned relation.

You need to think of this problem in terms of the Electric field? For a given potential difference, what is the field present? Or to think in the direction of the problem, what is the electric field required so that the necessary velocity is achieved?

 

[BuBbLeS01]

I think maybe it should be...
Change in K = q*Change in V

 

[rohanprabhu]

I think maybe it should be...
Change in K = q*Change in V

and yes.. you're right. Now, just substitute the values of 'm', 'v', 'q' and find what 'V' comes out to be.

also.. don't use the terminology 'change in V'. 'V' in itself refers to the potential difference.

Change in 'K', basically means the total Work done [following by 'Work-Energy theorem']. And 'V' is defined as the amount of work done per unit charge. So what you effectively did was first get the amount of work required to accelerate the given particle and then equate it with the amount of potential that could provide such energy [or in other words, do this work].

 

[BuBbLeS01]

Change in K = q*Change in V
(1/2 (6.6423x10^-24) * 26900000^2) / (1.6x10^-19) = V = 1.5x10^10 V
Is that right then?

 

[BuBbLeS01]

Oh wait to many 0's is velocity...answer I got is 1.5x10^8 V

 

[BuBbLeS01]

OMG...wrong number for mass lol...

(1/2 (6.6423x10^-27) * 2690000^2) / (1.6x10^-19) = V = 1.5x10^5 V
Is that right then?

 

[BuBbLeS01]

oh nooo its wrong :( I only have 1 try left...what did I do wrong?

 

[rohanprabhu]

OMG...wrong number for mass lol...

(1/2 (6.6423x10^-27) * 2690000^2) / (1.6x10^-19) = V = 1.5x10^5 V
Is that right then?

well.. seems right to me..

also.. try not to double post. Use the [edit] button in the bottom right corner to edit ur latest reply in case u have to add something. Make a new reply only when somebody has already commented on something or ur replies are far apart. Your last 3 replies were less than 5 mins. apart. Just keeps the forums clean. No offense.

EDIT:

i can't see anything wrong with the above. Are you sure it's a $He^+$ ion and not a $He^{+2}$ nucleus?