# Determine the momentum of the recoiling nucleus. Determine the angle to the nearest 1/100th degree.?

AN ISOTOPE EMITS AN ELECTRON AND AN ANTI-NEUTRINO DURING RADIOACTIVE DECAY. THE ELECTRON MOVES EAST WITH A MOMENTUM OF 9.28 X 10-26 KG∙ M/S AND THE ANTI-NEUTRINO MOVES NORTH WITH A MOMENTUM OF 7.47 X 10-27 KG∙ M/S AS SHOWN IN THE DIAGRAM BELOW.

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- NCSLv 74 months agoFavorite Answer
The combined momentum of the electron and anti-neutrino is

p = (9.28e-26, 7.47e-27) kg·m/s

using the (i, j) convention and assuming "i" is east.

This has magnitude

|p| = 9.31e-26 kg·m/s ◄

and so that's the magnitude of the momentum of the recoiling nucleus.

The direction is

Θ = 180º + arctan(7.47e-27/9.28e-26) = 185º ccw from "east"

or 4.60º S of West (below the -x axis)

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