A radioactive necleus at rest decays into a second necleus, an electron and a ne

Question

A radioactive necleus at rest decays into a second necleus, an electron and a neutrino. The electron and the neutrino are emitted at right angles and have a momenta of 9.10 * 10-23 kg * m/s, and 5.00 * 10-23 kg * m/s, respectively. What is the magnitude and direction of the momentum of the second (recoiling) necleus? (Hint: Since momentum is a vector, use conservation of momentum in two directions, independently.)

Magnitude

____________ kg * m/s

Direction

____________ degrees (measured from the direction opposite to the electron's momentum)

Explanation / Answer

let,

momentum of electron is P1=9*10^-23 kg*m/sec

momentum of neutrino is P2=5*10^-23 kg*m/sec

moementum of second necleus is P3

by law of conservation of momentum,

total momentum before decay=total momentum after decay

0=P1+P2+P3

and

P1 and P2 are at right angle each other,

P3=P3x+P3y

hence,

p3x=p1=9*10^-23 kg*m/sec

p3y=p2=5*10^-23 kg*m/sec

magnitude of P3=sqrt(P1^2+P2^2)

P3=sqrt((9*10^-23)^2+(5*10^-23)^2)

P3=1.03*10^-22 kg.m/sec

and

tan(theta)=P3y/P3x

tan(theta)=(5*10^-23)/(9*10^-23)

====>

theta=29.05 degrees,

angle between electron and nucleus is =180-29.05=150.95 degrees,

angle between neutrino and nucleus is =90+29.05=110.05 degrees,

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