Marie Curie first identified the element radium in 1898. She found that the radium-226 (mass 3.77x10^-25 kg) decays by emitting alpha particle (mass 6.64x10^-27 kg). If the alpha particle's speed is 2.4x10^6 m/s, what is the speed of the recoiling nucleus?
The mass of the recoiling nucleus (Polonium-222) is 3.70*10^-25 g.
Assume conservation of momentum for its speed, in the opposite direction from the alpha particle.
3.70*10^-25*V = 6.64*10^-27*2.4*10^6 m/s
Solve for V
V= 4.31 x 10^-46 ?
To find the speed of the recoiling nucleus, you can use the principle of conservation of momentum. According to this principle, the total momentum before an event is equal to the total momentum after the event, provided there are no external forces acting on the system.
In this case, we can consider the recoil of the nucleus as the event. Before the emission of the alpha particle, the system consists of the radium nucleus and the alpha particle. After the emission, the system consists of the recoiling nucleus and the emitted alpha particle.
The momentum of an object can be calculated by multiplying its mass by its velocity. So, we can write the conservation of momentum equation as:
(Momentum before) = (Momentum after)
The momentum before the emission is zero because the radium nucleus is initially at rest. The momentum after the emission is the sum of the momenta of the recoiling nucleus and the alpha particle.
Mass of the radium nucleus (m1) = 3.77x10^-25 kg
Velocity of the alpha particle (v1) = 2.4x10^6 m/s
Mass of the alpha particle (m2) = 6.64x10^-27 kg
Velocity of the recoiling nucleus (v2) = ?
We can use the conservation of momentum equation:
0 = (m1 x 0) + (m2 x v1) + (m2 x v2)
Since the initial momentum is zero, we can simplify the equation:
(m2 x v1) = -(m2 x v2)
Now, we can solve for the velocity of the recoiling nucleus (v2):
v2 = -(m2 x v1) / m2
Substituting the values into the equation:
v2 = -(6.64x10^-27 kg x 2.4x10^6 m/s) / 6.64x10^-27 kg
Performing the calculation:
v2 ≈ -2.4x10^6 m/s
Note: The negative sign indicates that the recoiling nucleus is moving in the opposite direction of the emitted alpha particle.