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 9.30 X 10^ -23 and 5.40 X 10^ -23 kg * m/s, respectively. What is the magnitude and direction of momentum of the second (recoiling) nucleus?
Im pretty sure the answer is 1.08kg*m/s x 10^-22 and the angle is 30 degrees, but can someone explain to me why you find the hypotnuse of the triangle and thats the magnitude of the second nuclues, and how do you find the angle. Explain in good detail please because i have no clue what i am doing
thanks

Explanation / Answer

Pxi = Pxf = Pelectron + Pnucleusx Pyi = Pyf = Pneutrino + Pnucleusy But the initial nucleus is at rest so Pxi = Pyi = 0 Then Pelectron = - Pnucleusx = - 9.3 x 10^-23 kg-m/s Pneutrino = - Pnucleusy = -5.4 x 10^-23 kg-m/s The magnitude |Pnucleus| = sqrt(Pnucleusx^2 + Pnucleusy^2) |Pnucleus| = 1.075 x 10^-22 kg-m/s Direction is q = 180 + atan(Pnucleusy/Pnucleusx) = 210.1 degress

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