Help with De Brogile Wavelength problem

[Whatlol]

[IMG]http://img440.imageshack.us/img440/5035/problemalphaparticles.png[/img]

 

I have managed to successfully completed both questions 8 a) and b) however, when i came to do c i got an answer twice as large.


I have a feeling this is because the charge on an alpha particle will be 2e (although not negative) can anyone confirm this, or provide the correct method to find the answer.

edit:  Here is how i tackled the question , and also the answer




I derived this formula to solve the problem, and then substituted the relevant figures in. Assuming that the charge should be 2e+ i.e two positive charges, then the problem is solved.

answer is 2.57x10^-5 V

[kenhung123]

Yea I think the charge is twice that of an electron.

[Whatlol]

Yea I think the charge is twice that of an electron.

Well a proton i suppose. but yea thats all i can think of.

[physics]

i ended up 0.000051 ev

[Whatlol]

i ended up 0.000051 ev

answer is 2.57x10^-5 V.... (its potential difference, not energy )
I got an answer close to yours initially. So only thing that  could be wrong is the charge on an alpha particle.


I derived this formula to solve the problem.

It worked for the previous questions, however when i attempted c) i got twice the answer.
So i multiplied the charge by two and it worked.

[physics]

well i did get the answer for the other 2 XD

[Whatlol]

well i did get the answer for the other 2 XD

awesome, but do you think that because the alpha particle has a charge of 2+, thats why the answer was half of the answer you got?