A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy of the proto

Question

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy of the proton and electron. The mass of a proton is 1836 times the mass of an electron. The mass of a proton is 1836 times the mass of an electron. What fraction of the total energy released goes into the kinetic energy of the proton?

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Xavia 5 months 2021-08-13T09:54:27+00:00 1 Answers 481 views 0

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  1. Dulcie
    0
    2021-08-13T09:56:11+00:00
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    Answer:

    5.44\times 10^-^4

    Explanation:

    Let A and B represent proton and electron respectively

    Total energy decay is:

    Q=K_A+K_B (1)

    where K=\frac{1}{2} mv^2 (2)

    The momentum of the particle is given by:

    \overrightarrow{p}=m\overrightarrow{v}

    After the decay we have \overrightarrow{p}_A+ \overrightarrow{p}_B=0 (3)

    Since the particles move in opposite direction,

    \overrightarrow{p}_A=m_Av_A, \overrightarrow{p}_B=-m_Bv_B

    From eq (3) we get m_Av_A=m_Bv_B, v_B=v_A\frac{m_A}{m_B} (4)

    From eq (2) we get K_B=\frac{1}{2} mv_B^2

    From eq (1) and (4), Q=K_A+\frac{1}{2}m_B(v_A \frac{m_A}{m_B})^2

    Q=K_A+\frac{1}{2}m_Av_A^2 (\frac{m_A}{m_B})

    K_A=\frac{1}{2}m_Av_A^2,Q=K_A+K_A(\frac{m_A}{m_B})=K_A(1+\frac{m_A}{m_B})

    m_A=1836m_B ,Q=K_A(1+1836)

    \frac{K_A}{Q}=\frac{1}{1837}=5.44\times 10^-^4

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