A uranium nucleus (mass 238 u, charge 92e) decays, emitting an alpha particle (mass 4 u, charge 2e) and leaving a thorium nucleus (mass 234 u, charge 90e). At the instant the alpha particle leaves the nucleus, the centers of the two are 5.5fm apart and essentially at rest.

Find their speeds when they're a great distance apart. Treat each particle as a spherical charge distribution.
Express your answers using two significant figures.

To find the speed of the alpha particle and the thorium nucleus when they are a great distance apart, we can make use of the principles of conservation of momentum and conservation of energy.

First, let's consider conservation of momentum. The initial momentum of the system (before alpha decay) is equal to the final momentum (after alpha decay). Since both the alpha particle and the thorium nucleus initially start at rest, the initial momentum is zero. The final momentum is given by the product of the mass and velocity of each particle.

Let's denote the speed of the alpha particle as v_alpha and the speed of the thorium nucleus as v_thorium. The mass of the alpha particle is 4u and the mass of the thorium nucleus is 234u.

So, the final momentum is:
(4u)(v_alpha) + (234u)(v_thorium) = 0

Now, let's consider conservation of energy. The initial energy of the system (before alpha decay) is equal to the final energy (after alpha decay). Since both particles start at rest, the initial energy is zero kinetic energy.

The final energy consists of the kinetic energy of the alpha particle and the kinetic energy of the thorium nucleus. The kinetic energy is given by 1/2 multiplied by the mass multiplied by the square of the velocity.

The final energy is:
(1/2)(4u)(v_alpha)^2 + (1/2)(234u)(v_thorium)^2 = 0

We can now solve these two equations simultaneously to find the values of v_alpha and v_thorium.

(4u)(v_alpha) + (234u)(v_thorium) = 0

(1/2)(4u)(v_alpha)^2 + (1/2)(234u)(v_thorium)^2 = 0

Let's solve these equations:

From the first equation, we can express v_alpha in terms of v_thorium:
v_alpha = -[(234u)(v_thorium)] / (4u)

Now we substitute this value of v_alpha into the second equation:
(1/2)(4u)[-(234u)(v_thorium) / (4u)]^2 + (1/2)(234u)(v_thorium)^2 = 0

Simplifying this equation:
(1/2)(4u)(234u)^2(v_thorium^2) / (16u^2) + (1/2)(234u)(v_thorium)^2 = 0

Now we solve for v_thorium:

Simplifying:
(234u^2)(v_thorium^2) / (8u) + (234u)(v_thorium)^2 = 0

Combining like terms:
(234u)(v_thorium^2)[(u/8) + 1] = 0

Now we have two possibilities:
1. (234u)(v_thorium^2) = 0
2. (u/8) + 1 = 0

Since u is nonzero, (234u)(v_thorium^2) cannot be zero. Therefore, we consider the second possibility:

(u/8) + 1 = 0

Solving for u:
u = -8

However, since mass cannot be negative, this possibility is not valid.

Therefore, we cannot determine the values of v_alpha and v_thorium using the given information.