A uranium nucleus (mass 238 u), initially at rest, undergoes radioactive decay. After an alpha particle (mass 4.0 u) is emitted, the remaining nucleus is thorium (mass 234 u). If the alpha particle is moving at 0.052 the speed of light, what is the recoil speed of the thorium nucleus?
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To find the recoil speed of the thorium nucleus after the alpha particle is emitted, we can use the conservation of momentum.
1. First, let's calculate the initial momentum of the uranium nucleus. Since it is initially at rest, its momentum is zero.
2. Next, let's calculate the momentum of the alpha particle. The momentum (p) is given by the formula:
p = mv
where m is the mass of the alpha particle and v is its velocity. The mass of the alpha particle is 4.0 u, and its velocity is 0.052 times the speed of light.
Thus, the momentum of the alpha particle is:
p_alpha = (4.0 u) * (0.052 c)
= 0.208 u*c
(Note: u represents the atomic mass unit, and c represents the speed of light)
3. According to the conservation of momentum, the total momentum before and after the decay must be zero. Therefore, the momentum of the thorium nucleus after the alpha particle is emitted will be equal in magnitude but opposite in direction to the momentum of the alpha particle.
Let's denote the recoil speed of the thorium nucleus as v_thorium and its mass as M_thorium.
Then, the momentum of the thorium nucleus is:
p_thorium = - p_alpha
where the negative sign indicates the opposite direction.
4. We can use the formula for momentum:
p = mv
Applying it to the thorium nucleus, we have:
(- p_alpha) = (M_thorium) * (v_thorium)
Rearranging the equation, we find:
v_thorium = - (p_alpha) / (M_thorium)
5. Finally, substitute the values into the equation to calculate the recoil speed of the thorium nucleus:
v_thorium = - (0.208 u*c) / (234 u)
= - 0.000888 c
(Note: The negative sign indicates that the thorium nucleus moves in the opposite direction to the alpha particle.)
Therefore, the recoil speed of the thorium nucleus is approximately -0.000888 times the speed of light.