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.(adsbygoogle = window.adsbygoogle || []).push({});

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 JIs 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Alpha Decay and Momentum

**Physics Forums | Science Articles, Homework Help, Discussion**