A uranium nucleus 238U may stay in one piece for billions of years, but sooner or later it decays into an a 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 a particle is emitted at a speed of 1.07×107 m/s, what would be the recoil speed of the thorium nucleus?
Answer in units of m/s.
I got -183113.4021m/s, but that was wrong, and so was 183113.4021m/s
To find the recoil speed of the thorium nucleus, we can make use of the principle of conservation of momentum.
First, let's calculate the initial momentum of the system, which consists of the uranium nucleus alone. Since the uranium nucleus is at rest initially, its momentum is zero.
Next, let's calculate the final momentum of the system, which consists of the alpha particle (a) and the thorium nucleus (Th) after the decay. The total momentum of the system must also be zero, as there are no external forces acting on it.
The momentum of the alpha particle (p₁) can be calculated using the equation p = mv, where m is the mass and v is the speed:
p₁ = (mass of alpha particle) × (speed of alpha particle)
p₁ = (6.64×10−27 kg) × (1.07×107 m/s)
The momentum of the thorium nucleus (p₂) can be calculated in a similar way:
p₂ = (mass of thorium nucleus) × (recoil speed of thorium nucleus)
Since the sum of the momenta before and after the decay is zero, we can write the equation:
0 = p₁ + p₂
Rearranging this equation, we can solve for the recoil speed of the thorium nucleus:
recoil speed of thorium nucleus = - p₁ / (mass of thorium nucleus)
Let's plug in the values and calculate:
recoil speed of thorium nucleus = - [(6.64×10−27 kg) × (1.07×107 m/s)] / (3.88 × 10−25 kg)
Calculating this expression, we get:
recoil speed of thorium nucleus ≈ -1825.77 m/s
Rounding to the appropriate number of significant figures, the recoil speed of the thorium nucleus would be approximately -1825.8 m/s.
To find the recoil speed of the thorium nucleus, we can use the principle of conservation of momentum.
The initial momentum of the system is zero since the uranium nucleus is at rest. Therefore, the final momentum of the system must also be zero to satisfy the conservation of momentum.
Let's denote the recoil speed of the thorium nucleus as v. The momentum of the alpha particle is given by p_alpha = m_alpha * v_alpha, and the momentum of the thorium nucleus is given by p_thorium = m_thorium * v.
Since the final momentum is zero, we have:
p_alpha + p_thorium = 0
Substituting in the given masses and speeds:
(m_alpha * v_alpha) + (m_thorium * v) = 0
Now, let's plug in the values:
(6.64 × 10^-27 kg) * (1.07 × 10^7 m/s) + (3.88 × 10^-25 kg) * v = 0
Solving for v:
(6.64 × 10^-27 kg) * (1.07 × 10^7 m/s) = -(3.88 × 10^-25 kg) * v
v = - (6.64 × 10^-27 kg) * (1.07 × 10^7 m/s) / (3.88 × 10^-25 kg)
v ≈ -1.82 × 10^5 m/s
Note that the negative sign indicates that the thorium nucleus recoils in the opposite direction to the alpha particle emission.
Therefore, the recoil speed of the thorium nucleus is approximately 1.82 × 10^5 m/s.
Massparticle*velocityParticle=massU234*V
solve for V.
6.46E-27*1.07E7=3.88E-25*V
I agree, your answers are not correct.