A radioactive nucleus at rest decays into a second nucleus, an electron, and a n

Question

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 pe = 9.67×1023 kgm/s and p = 6.36×1023 kgm/s , respectively. (Figure 1)

Part A

Determine the magnitude of the momentum of the second (recoiling) nucleus. Express your answer to three significant figures and include the appropriate units. pnuc = ?

Part B

Determine the angle between the momentum of the electron and the momentum of the second (recoiling) nucleus. Express your answer using three significant figures.

Explanation / Answer

(a)

Since the original nucleus was at rest its momentum (p) = 0. Since momentum is conserved the momentum of the three particles will sum to zero also.

So p (nucleus) + p (electron) + p(neutrino) = 0 (Vectors)

So p (nucleus) = - p(electron) - p( neutrino) Let the direction of the electron be in the - x direction and that of the neutrino be + y

So the direction of the nucleus is in the +x And -y directions and is equal to

9.67x10^-23 kg-m/s (+x) and 6.36x10^-23 kg-m/s (-y )

Therefore the magnitude of p (nucleus)

= sqrt((9.67x10^-23)^2 +(6.36x10^-23)^2)

= 1.16x10^-22kg-m/s

(b)

The angle

= arctan (6.36x10^-23/9.67x10^-23)

= 33.3 degrees

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