Alpha Decay and Momentum

[Soaring Crane]

A 232 Th (thorium) nucleus at rest decays to a 228 Ra (radon) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is 6.54 x 10^-13 J. An alpha particle has 1.76% of the mass of a 228 Ra nucleus.

a) Calculate the kinetic energy of the recoiling 228 Ra nucleus.

b) Calculate the kinetic energy of the alpha particle.


Please check to see if my setup is correct. If it is not, kindly tell me which setup(s) I am supposed to use.

I know Conservation of Momentum is involved in this explosion.

Momentum is 0 since Th nucleus is at rest before alpha decay.

Therefore, m_alpha*v_alpha = -m_Ra*v_Ra, so .0176*m_Ra*v_alpha = -m_Ra*v_Ra.

m_Ra cancels:

.0176*v_alpha = -v_Ra

From kinetic energy,

KE = (0.5)*m_Ra*(v_Ra)^2 + (0.5)*0.0176*m_Ra*(v_alpha)^2, where KE = is 6.54 x 10^-13 J Is this right?

v_alpha = (-v_Ra)/.0176

Substitue this in KE formula.

KE = (0.5)*m_Ra*(v_Ra)^2 + (0.5)*0.0176*m_Ra*[(-v_Ra)/.0176)]^2

KE = (0.5)*m_Ra*(v_Ra)^2 + (0.5)*m_Ra*(v_Ra)^2*(1/.0176)

Now I used mass_Radon nucleus = 3.6*10^-25 kg to solve for v_Ra, but this is not given. Are the masses supposed to cancel?

Solving for v_Ra,

v_Ra = sqrt[(3.633*10^12)/57.81818)] = 250680.3047 m/s

Individual KE:

KE_Ra = (3.6*10^-25 kg)*(250680.3047 m/s)^2*0.5 = 1.13*10^-14 J

KE_alpha = (3.6*10^-25 kg)*(-250680.3047 m/s)^2*0.5*(1/.0176) = 6.43*10^-13 J

I would appreciate any help.

Thank you.

[Office_Shredder]

It was a bit tough to read through, but it looked right through when you solved for the velocity.

A nice way to make things look a lot neater, instead of writing KE_alpha, writing KEalpha doesn't even require using latex, you just use [.sub] and [./sub] tags without the '.' You can also do the same thing with [.sup] and [./sup] to get super scripts, like this: v1/2

[Soaring Crane]

Thanks. I will try to use the tags in my future posts.

Was it correct of me to use mass_Radon nucleus = 3.6*10^-25 kg even though it was not given?