A U-238 nucleus originally at rest , decays by emitting an alpha- particle , say with a velocity of vm/s. The recoil (in m/s) of the residual nucleus is.?
Ans: 4v/238
the mass of the recoiling nucleus is
... 238 - 4
r * 234 = v * 4
work this out with conservation of momentum, the atomic mass of the alpha is 4amu
To determine the recoil velocity of the residual nucleus after an alpha particle is emitted, we can apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum before an event is equal to the total momentum after the event, assuming no external forces act on the system. In this case, we are considering a closed system where only the U-238 nucleus and the alpha particle are involved.
Initially, the U-238 nucleus is at rest, so its momentum is zero. The alpha particle, with a mass of four atomic mass units (amu), is emitted with a velocity of vm m/s.
After the alpha decay, the residual nucleus is left with a recoil velocity, which we need to determine. Let's assume its velocity is v_r m/s.
Using the conservation of momentum, we can write:
Momentum before decay = Momentum after decay
(Initial momentum of U-238 nucleus) + (Initial momentum of alpha particle) = (Final momentum of residual nucleus)
Since the U-238 nucleus is initially at rest, its momentum is zero:
0 + (mass of alpha particle) × (velocity of alpha particle) = (mass of residual nucleus) × (velocity of residual nucleus)
(4 amu) × (vm m/s) = (mass of residual nucleus) × (v_r m/s)
Now, we need to relate the masses of the particles involved. The mass of the alpha particle (m_alpha) is four times the mass of a proton (mp):
m_alpha = 4 × mp
The mass of the residual nucleus (m_residual) is the mass of the original U-238 nucleus (m_U238) minus the mass of the alpha particle:
m_residual = m_U238 - m_alpha
Finally, we can substitute these expressions into our momentum conservation equation:
(4 amu) × (vm m/s) = (m_U238 - (4 × mp)) × (v_r m/s)
Rearranging the equation to solve for the recoil velocity (v_r):
v_r = (4 amu × vm m/s) / (m_U238 - (4 × mp))
Since the question asks for the result in terms of v (vm) and the mass of the U-238 nucleus (m_U238), we substitute the mass of a proton (mp) with 1 amu:
v_r = (4 × vm m/s) / (m_U238 - (4 amu))
Considering the mass of the U-238 nucleus is 238 amu:
v_r = (4 × vm m/s) / (238 amu - (4 amu))
Simplifying:
v_r = (4 × vm m/s) / 238 amu
Therefore, the recoil velocity of the residual nucleus is v_r = (4v) / 238.