An atomic nucleus of radon initially moving at 530 m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 410 m/s. If the alpha particle has a mass of 4u, and the original nucleus of radon had a mass of 222u, what speed does the alpha particle have when it is emitted?
u is the SI unit for atomic mass. Even though it is not in kg, you don't need to do the conversion since the units of mass will cancel.
222 * 530 = original momentum
4 v + 218 * 410 = final momentum
original momentum = final momentum
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after the emission of the alpha particle should be the same.
The momentum (p) of an object can be calculated as the product of its mass (m) and velocity (v): p = m * v.
Initially, the momentum of the radon nucleus is given as:
p_initial = m_initial * v_initial = 222u * 530 m/s
After the emission of the alpha particle, the momentum of the new nucleus is given as:
p_final = m_final * v_final = (222u - 4u) * 410 m/s
Since the total momentum is conserved, we can set p_initial equal to p_final and solve for v_final:
m_initial * v_initial = (m_final - m_alpha) * v_final
222u * 530 m/s = (222u - 4u) * 410 m/s
Simplifying the equation, we have:
530 m/s * 222u = 410 m/s * 218u
u cancels out on both sides of the equation and we are left with:
530 * 222 = v_final * 410
Solving for v_final, we have:
v_final = (530 * 222) / 410
Calculating the value:
v_final = 286.10 m/s
Therefore, the speed of the alpha particle when it is emitted is approximately 286.10 m/s.