A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of 9.15×10−23 kg . m/s and 5.19×10−23 kg . m/s, respectively.
1.)What is the magnitude of the momentum of the second (recoiling) nucleus?
2.)What are the direction of the momentum of the second (recoiling) nucleus? (theta = from the momentum of the electron)
To find the magnitude of the momentum of the second (recoiling) nucleus, we can use the principle of conservation of momentum. Since the total momentum before and after the decay must be conserved, we can write:
(momentum of nucleus before decay) = (momentum of nucleus after decay) + (momentum of electron) + (momentum of neutrino)
Since the radioactive nucleus is initially at rest, its momentum before the decay is zero. Therefore, we can rewrite the equation as:
0 = (momentum of nucleus after decay) + (momentum of electron) + (momentum of neutrino)
Now we can solve for the magnitude of the momentum of the second nucleus:
(magnitude of momentum of second nucleus) = (magnitude of momentum of electron) + (magnitude of momentum of neutrino)
Given the momentum of the electron is 9.15×10^−23 kg.m/s and the momentum of the neutrino is 5.19×10^−23 kg.m/s, we can substitute these values into the equation:
(magnitude of momentum of second nucleus) = 9.15×10^−23 kg.m/s + 5.19×10^−23 kg.m/s
= 14.34×10^−23 kg.m/s
= 1.434×10^−22 kg.m/s
So, the magnitude of the momentum of the second (recoiling) nucleus is 1.434×10^−22 kg.m/s.
For the direction of the momentum of the second (recoiling) nucleus, we are given that it is emitted at right angles to the momentum of the electron. This means that the direction of the momentum of the nucleus and the electron are perpendicular to each other.
However, without any specific information about the angle theta provided in the question, we cannot determine the exact direction of the momentum of the second nucleus.
To find the magnitude of momentum of the second (recoiling) nucleus, we can use the principle of conservation of momentum. According to this principle, the total momentum before the decay is equal to the total momentum after the decay.
1.) The total initial momentum is zero since the radioactive nucleus is at rest. Therefore, the total final momentum must also be zero.
Let's denote the magnitude of the momentum of the recoiling nucleus as P2, and the angle between the momentum of the electron and the recoiling nucleus as θ.
Using vector addition, we can break down the momentum of the electron into its x-component (Px) and y-component (Py):
Px = 9.15×10^−23 kg . m/s * cosθ
Py = 9.15×10^−23 kg . m/s * sinθ
Similarly, we can break down the momentum of the neutrino into its x-component (Pn) and y-component (Pn):
Pn = 5.19×10^−23 kg . m/s * cos(90° - θ) = 5.19×10^−23 kg . m/s * sinθ
Pn = 5.19×10^−23 kg . m/s * cosθ
Since the total final momentum is zero, we have:
P2x + Px + Pn = 0
P2y + Py + Pn = 0
Substituting the values we have:
P2x + 9.15×10^−23 kg . m/s * cosθ + 5.19×10^−23 kg . m/s * cosθ = 0
P2y + 9.15×10^−23 kg . m/s * sinθ + 5.19×10^−23 kg . m/s * sinθ = 0
Simplifying the equations:
P2x = -14.34×10^−23 kg . m/s * cosθ
P2y = -14.34×10^−23 kg . m/s * sinθ
The magnitude of the momentum of the recoiling nucleus, P2, can be determined using the Pythagorean theorem:
P2 = sqrt(P2x^2 + P2y^2)
2.) To find the direction of the momentum of the recoiling nucleus, we need to determine the angle, θ, made by the momentum of the recoiling nucleus with the momentum of the electron.
The direction can then be expressed as the angle θ from the momentum of the electron, as given in the question.
Therefore, the magnitude of the momentum of the second (recoiling) nucleus is P2 = sqrt(P2x^2 + P2y^2), and the direction of the momentum is θ from the momentum of the electron.
Draw yourself a triangle with the neutrino and electron forming two sides at right angle. The hypotenuse will be the momentum of the nucleus, so that the sum of momenta remains zero.
Solve the triangle using simple rules of trigonometry.
The angle theta is, I believe,
180 - arctan 5.19/9.15
= 150.4 degrees