A uranium nucleus 238U may stay in one piece
for billions of years, but sooner or later it de-
cays into an � particle of mass 6.64×10−27 kg
and 234Th nucleus of mass 3.88 × 10−25 kg,
and the decay process itself is extremely fast
(it takes about 10−20 s). Suppose the uranium
nucleus was at rest just before the decay.
If the � particle is emitted at a speed of
5 × 106 m/s, what would be the recoil speed
of the thorium nucleus?
Answer in units of m/s
m1 = 6.64•10^−27 kg , v1 =5•10^6 m/s,
m2 = 3.88•10^-25 kg, v2 = ?
0 = m1•v1 – m1•v2,
v2 = m1•v1/v2 =
= 6.64•10^-27•5•10^6/3.88•10^-25 =
= 8.56•10^4 m/s.
To find the recoil speed of the thorium nucleus, we can use the law of conservation of momentum. According to this law, the total momentum before the decay should be equal to the total momentum after the decay.
Before the decay, the uranium nucleus is at rest, so its momentum is zero. The momentum of an object is given by the product of its mass and velocity, so the initial momentum is 0 kg.m/s.
After the decay, there are two particles: the α particle (α) and the thorium nucleus (Th). Let's assume the recoil velocity of the thorium nucleus is v.
The total momentum after the decay can be calculated as the sum of the momenta of the α particle and the thorium nucleus. Since they are moving in opposite directions, the momentum of the α particle is positive, and the momentum of the thorium nucleus is negative.
The momentum of the α particle (pα) can be calculated using the equation: pα = mα * vα, where mα is the mass of the α particle and vα is its velocity.
The momentum of the thorium nucleus (pTh) can be calculated using the equation: pTh = mTh * vTh, where mTh is the mass of the thorium nucleus and vTh is its recoil velocity.
According to the law of conservation of momentum, the sum of the momenta before the decay (0) should be equal to the sum of the momenta after the decay:
0 = pα + pTh
Substituting the values into the equation, we can solve for the recoil velocity of the thorium nucleus (vTh):
0 = (mα * vα) + (mTh * vTh)
Since the α particle is emitted at a velocity of 5 × 10^6 m/s, we can substitute this value into the equation:
0 = (mα * 5 × 10^6) + (mTh * vTh)
Solving for vTh, we can isolate the variable:
vTh = - (mα * 5 × 10^6) / mTh
Now we can substitute the values for the mass of the α particle (mα = 6.64 × 10^(-27) kg) and the mass of the thorium nucleus (mTh = 3.88 × 10^(-25) kg) into the equation to find the recoil velocity:
vTh = - (6.64 × 10^(-27) kg * 5 × 10^6 m/s) / (3.88 × 10^(-25) kg)
Calculating this expression will give us the recoil velocity of the thorium nucleus in units of m/s.