Chapter 7, Problem 40P

To determine

To calculate: The magnitude and direction of the momentum of the second (recoiling) nucleus.

Answer to Problem 40P

  $P_{nu}=10.75\times {10}^{-23}$ kg m/sec

  $\theta =30.11°$ from North to West.

Explanation of Solution

Given:

A radioactive nucleus is initially at rest and then decays into a nucleus consisting of electron and neutrino. The electron and neutrino are emitted at right angles and their momentum is $9.30\times {10}^{-23}$ kg m/sec and $5.40\times {10}^{-23}$ kg m/sec.

Formula used:

  $P_{nu}{}^2=P_e{}^2+P_n{}^2$ Here,

P_nu is the momentum of the nucleus.

P_e is the momentum of the electron.

P_n is the momentum of the neutrino

Calculation:

Electrons and neutrino are at right angles. Nucleus is emitted at opposite to the right angle. So, the three P_{nu ,} P_e and P_n are momentum vectorsFrom $P_{nu}{}^2=P_e{}^2+P_n{}^2$

  $P_{nu}{}^2={(9.30\times {10}^{-23})}^2+{(5.40\times {10}^{-23})}^2$

  $=86.49\times {10}^{-46}+29.16\times {10}^{-46}$

  $=115.65\times {10}^{-46}$

  $P_{nu}=\sqrt{115.65\times {10}^{-46}}$

  $P_{nu}=10.75\times {10}^{-23}$ kg m/sec

Direction is $\tan \theta =\frac{P_n}{P_e}=\frac{5.40\times {10}^{-23}}{9.30\times {10}^{-23}}=\frac{5.4}{9.3}$

  $\theta ={\tan }^{-1}\left(\frac{5.4}{9.3}\right)$

  $\theta =30.11°$ from North to West.