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 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

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