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 multiplied by 10-23 kg·m/s, and 5.00 multiplied by 10-23 kg·m/s, respectively. What is the magnitude and direction of the momentum of the second (recoiling) nucleus?(Hint: Since momentum is a vector, use conservation of momentum in the two directions, independently.)
Magnitude
kg·m/s
Direction
° (measured from the direction opposite to the electron's momentum)
Explanation / Answer
According to law of conservation of linear momentum , the net momentun acting on the system is zero . Hence the rocoil velocity of the second nucleus is equals to the net mometum of the electron and neutrino Let momentum of the electron Pe = 9.3*10-23 kg .m/s is along + ve x -direction momentum of the neutron Pn = 5*10-23 kg .m/s is along +ve y direction The mometum of the nucleus is 9.3*10-23 kg .m/s is along - ve x -direction , 5*10-23 kg .m/s is along - ve y direction momentum of the neutron Pn = 5*10-23 kg .m/s is along +ve y direction The mometum of the nucleus is 9.3*10-23 kg .m/s is along - ve x -direction , 5*10-23 kg .m/s is along - ve y direction Pnet = sqrt ( ( Pe )2 + ( Pn )2 ) Pnet =10.55 *10-23 kg .m/s direction = tan ( Pn / Pe ) = 28.260 below the - X axis (OR) = 208.26 0 cunterclock wise direction from the +ve X axis