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 2.20E+05 m/s? Assume the recoiling nucleus has a mass 54 times as great as that of the alpha particle.
54M*V=M*2.2E5
V=2.2E5/54
To determine the speed of the recoiling nucleus, we can use the conservation of momentum.
The momentum is conserved before and after the decay, so we can write:
(mass of the alpha particle)(speed of the alpha particle) + (mass of the recoiling nucleus)(speed of the recoiling nucleus) = 0
Since the mass of the recoiling nucleus is 54 times as great as that of the alpha particle, we can rewrite the equation as:
(1)(2.20E+05 m/s) + (54)(speed of the recoiling nucleus) = 0
Simplifying the equation gives us:
2.20E+05 + 54(speed of the recoiling nucleus) = 0
54(speed of the recoiling nucleus) = -2.20E+05
Dividing both sides of the equation by 54 gives us the speed of the recoiling nucleus:
speed of the recoiling nucleus = (-2.20E+05) / 54
Calculating this gives us:
speed of the recoiling nucleus ≈ -4074 m/s
Therefore, the speed of the recoiling nucleus will be approximately -4074 m/s.
To find the speed of the recoiling nucleus, we can use the principle of conservation of momentum. This principle states that the total momentum of an isolated system remains constant before and after any interaction.
Let's denote the initial momentum of the system as p_initial and the final momentum as p_final.
Before the decay, the total momentum of the system is given by:
p_initial = m_alpha * v_alpha
Where:
m_alpha = mass of the alpha particle
v_alpha = velocity of the alpha particle
After the decay, the alpha particle is moving in one direction, and the recoiling nucleus is moving in the opposite direction. The masses and velocities of the recoiling nucleus and alpha particle will be denoted as m_nucleus, v_nucleus, and m_alpha, v_alpha, respectively.
The final momentum of the system is given by:
p_final = m_alpha * v_alpha + m_nucleus * v_nucleus
Since we know the mass of the recoiling nucleus and that it is 54 times greater than the alpha particle, we can write:
m_nucleus = 54 * m_alpha
Now, since the system is isolated, the momentum before and after the decay must be equal:
p_initial = p_final
Therefore,
m_alpha * v_alpha = m_alpha * v_alpha + 54 * m_alpha * v_nucleus
Now, we can solve this equation to find the velocity of the recoiling nucleus (v_nucleus):
0 = m_alpha * v_alpha - 54 * m_alpha * v_nucleus
Rearranging and simplifying,
54 * m_alpha * v_nucleus = m_alpha * v_alpha
Dividing both sides by 54 * m_alpha,
v_nucleus = v_alpha / 54
Now we can substitute the given value for the velocity of the alpha particle to find the velocity of the recoiling nucleus:
v_nucleus = 2.20E+05 m/s / 54
Calculating this:
v_nucleus = 4074.07 m/s
Therefore, the speed of the recoiling nucleus is approximately 4074.07 m/s.