Question

A radioactive nucleus at rest decays into a second nucleus, electron, and neutrino. The electron and neutrino are emitted at right angles. The electron has a momenta of 9.00x10^-23 kg m/s and the neutrino has a momenta of 6.65x10^-23 kg m/s.

Determine the direction of the momentum of the second (recoiling) nucleus. Assume that neutrino moves along -x axis and the electron moves along the -y axis.

**The answer of theta needs to be the degrees
counterclockwise from the +x axis.**

Thank you in advance and if right, I will rate and up the answer.

Answer #1

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)

Given the direction of the electron is in - Y direction and that of the neutrino is - X

===>So the direction of the nucleus is in the +X and +Y directions and momentum is equal to:

9.00x10^-23 kg-m/s (+Y) and 6.65x10^-23 kg-m/s (+X )

**Therefore the magnitude of p (nucleus)**

= sqrt((9.00x10^-23)^2 +(6.65x10^-23)^2)

**= 1.119x10^-22 kgm/s (ANSWER ===>
Magnitude)**

**The Direction (angle)**

= arctan (9.00x10^-23 / 6.65x10^-23)

= arctan (1.3533)

**= 53.54° (ANSWER===>DIRECTION)**

**Kindly do like if it helped!**

**Thank you!!!**

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