An atomic nucleus at rest decays radioactively into an alpha particle and a nucleus smaller than the original. What will be the speed of this recoiling nucleus if the speed of the alpha particle is 1.20E+05 m/s? Assume the recoiling nucleus has a mass 56 times as great as that of the alpha particle.
momentum=zero
Malpha*1.20E05=Malpha*56*v
solve for v.
To find the speed of the recoiling nucleus, we can use the principle of conservation of momentum. In a closed system, the total momentum before the decay should be equal to the total momentum after the decay.
Let's denote the mass of the alpha particle as m₁ and its velocity as v₁, and the mass of the recoiling nucleus as m₂ and its velocity as v₂.
Before the decay:
The momentum of the alpha particle is given by:
p₁ = m₁ * v₁
After the decay:
The momentum of the alpha particle is still given by:
p₁' = m₁ * v₁
The momentum of the recoiling nucleus is given by:
p₂ = m₂ * v₂
According to the conservation of momentum:
p₁ + p₂ = p₁' + p₂
Substituting in the given values:
m₁ * v₁ + m₂ * v₂ = m₁ * v₁ + p₂
Since the alpha particle is at rest before the decay, its initial momentum (m₁ * v₁) is zero. Therefore, we can simplify the equation to:
m₂ * v₂ = p₂
Now, considering the mass relation given in the question:
m₂ = 56 * m₁
Substituting this value into the equation, we get:
56 * m₁ * v₂ = p₂
Given that the velocity of the alpha particle (v₁) is 1.20E+05 m/s, we can substitute the values and solve for v₂:
56 * (1.20E+05 m/s) = v₂
Calculating this expression, we find:
v₂ ≈ 6.72E+06 m/s
Therefore, the speed of the recoiling nucleus will be approximately 6.72E+06 m/s.