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 9.40 x 10^-23 kg·m/s, and 6.00 x 10^-23 kg·m/s, respectively. What is the magnitude and direction of the momentum of the second (recoiling) nucleus?

To determine the magnitude and direction of the momentum of the second nucleus, we can apply the law of conservation of momentum. According to this law, the total momentum before the decay should be equal to the total momentum after the decay.

Let's denote the momentum of the second nucleus as p_nucleus2 and assume it has a mass m_nucleus2. We can break down the given momenta of the electron and neutrino into their x and y components:

For the electron:
p_electron = (9.40 x 10^-23 kg·m/s) in the x-direction
+ (0 kg·m/s) in the y-direction (since they are emitted perpendicular to each other)

For the neutrino:
p_neutrino = (0 kg·m/s) in the x-direction
+ (6.00 x 10^-23 kg·m/s) in the y-direction

According to the conservation of momentum:
0 kg·m/s = p_electron_x + p_neutrino_x = (9.40 x 10^-23) kg·m/s
p_nucleus2 = -p_electron_y - p_neutrino_y = - (9.40 x 10^-23 + 6.00 x 10^-23) kg·m/s

Therefore, the magnitude of the momentum of the second nucleus is:
|p_nucleus2| = |-(9.40 x 10^-23 + 6.00 x 10^-23)| kg·m/s = (15.40 x 10^-23) kg·m/s

To find the direction of the momentum of the second nucleus, we need to consider the signs of the electron and neutrino momenta, as well as their orientations.

Given that the electron is emitted in the positive x-direction and the neutrino is emitted in the positive y-direction, we can conclude that the second nucleus must be emitted in the negative y-direction to conserve momentum.

Therefore, the magnitude and direction (angle with respect to the positive x-axis) of the momentum of the second nucleus are:
Magnitude: 15.40 x 10^-23 kg·m/s
Direction: 180° (in the negative y-direction)

To solve this problem, we will 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.

Let's assume the initial momentum of the radioactive nucleus is zero since it is at rest. Therefore, the total momentum before the decay is just the momentum of the electron and neutrino.

P_initial = P_electron + P_neutrino

We are given the momenta of the electron and neutrino:

P_electron = 9.40 x 10^-23 kg·m/s

P_neutrino = 6.00 x 10^-23 kg·m/s

Thus, the total initial momentum before the decay is:

P_initial = (9.40 x 10^-23 kg·m/s) + (6.00 x 10^-23 kg·m/s)

P_initial = 15.40 x 10^-23 kg·m/s

According to the principle of conservation of momentum, the total momentum after the decay is also equal to P_initial.

Now, since the electron and neutrino are emitted at right angles, their momenta can be added as vectors using the Pythagorean theorem:

P_recoiling nucleus = √(P_initial^2 - P_electron^2 - P_neutrino^2)

P_recoiling nucleus = √((15.40 x 10^-23 kg·m/s)^2 - (9.40 x 10^-23 kg·m/s)^2 - (6.00 x 10^-23 kg·m/s)^2)

P_recoiling nucleus = √(238.81 x 10^-46 kg^2·m^2/s^2 - 88.36 x 10^-46 kg^2·m^2/s^2 - 36 x 10^-46 kg^2·m^2/s^2)

P_recoiling nucleus = √(114.45 x 10^-46 kg^2·m^2/s^2)

P_recoiling nucleus = 10.70 x 10^-23 kg·m/s

So, the magnitude of the momentum of the second (recoiling) nucleus is 10.70 x 10^-23 kg·m/s.

Since the electron and neutrino are emitted at right angles, we can consider their momenta as perpendicular components of the total momentum after the decay. Thus, the direction of the momentum of the second nucleus is perpendicular to the plane of the emission of the electron and neutrino. The exact direction cannot be determined without further information.

Therefore, the magnitude of the momentum of the second (recoiling) nucleus is 10.70 x 10^-23 kg·m/s, and its direction is perpendicular to the plane of the emission of the electron and neutrino.