The Brogile wavelength of an Alpha particle

[mrausum]

1. The problem statement, all variables and given/known data

An alpha particle of energy 5.78 MeV is emitted from a particular nucleus.
Calculate the de Broglie wavelength of the alpha particle. How does it compare
with the nuclear diameter, which is known to be approximately 2x10−14 m.


2. Relevant equations

E=h/p and E=hf

3. The attempt at a solution

I was thinking of using E=hc/lamda but this would be wrong because alpha particles don't travel at c.

[rock.freak667]

The energy of 5.78MeV is entirely kinetic.

[mrausum]

so how do i do the question?

[rock.freak667]

so how do i do the question?

Use Ek=1/2mv2

[mrausum]

well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?

[rock.freak667]

well it hasn't given the mass. But are you saying work out v from E=1/2mv^2, then f from E/h. Then wavelength from lamda=v/f?

find v, then use

\lambda = \frac{h}{mv}

I am sure you can look up the mass of an alpha particle. I am not sure how else you can solve this problem without my suggestion though.

[HjGanap]

how do you get lambda = h/mv
i know E = hc/lambda = h/mv
so, if i combine both equations, shouldnt i suppose to get lambda = cmv ?
did i miss something or what.

[mrausum]

how do you get lambda = h/mv
i know E = hc/lambda = h/mv
so, if i combine both equations, shouldnt i suppose to get lambda = cmv ?
did i miss something or what.

p = h/lamda
lamda = h/p=h/mv

You can't use E = hc/lamda because alpha particles don't travel at the speed of light.