An atomic nucleus of radon initially moving at 515 m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 417 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?
Use conservation of momentum
222*515 = 417*(222-4) + 4*valpha
where valpha is the speed of the alpha particle
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the emission of the alpha particle must be equal to the total momentum after the emission.
The momentum of an object is given by the product of its mass and velocity. So, let's first calculate the momentum of the original radon nucleus before emission.
Momentum of the original nucleus = mass of the nucleus x velocity of the nucleus
Mass of the original nucleus = 222u
Velocity of the original nucleus = 515 m/s
Momentum of the original nucleus = 222u x 515 m/s
Next, let's calculate the momentum of the new radon nucleus after the emission.
Momentum of the new nucleus = mass of the nucleus x velocity of the nucleus
Mass of the new nucleus = mass of the original nucleus - mass of the alpha particle
= 222u - 4u
= 218u
Velocity of the new nucleus = 417 m/s
Momentum of the new nucleus = 218u x 417 m/s
According to the principle of conservation of momentum, these two momenta must be equal. Therefore, we can write the following equation:
Momentum of the original nucleus = Momentum of the new nucleus
222u x 515 m/s = 218u x 417 m/s
Now, let's solve this equation to find the missing velocity of the alpha particle.
(222u x 515 m/s) / 218u = velocity of the alpha particle
Using these numbers in the equation above, we can calculate the velocity of the alpha particle when it is emitted. This gives us the answer to your question.