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.

1. What fraction of the total energy released goes into the kinetic energy of the proton?

Express your answer as a percentage.

Kp/Ktot= _______________________ %
I have already tried these answers and they’re incorrect

-1/1836

-1/1837

-1873

-99

-5.44×10^(-4)

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Khoii Minh 5 years 2021-08-20T16:59:00+00:00 1 Answers 82 views 0

Answers ( )

  1. minhkhang
    0
    2021-08-20T17:00:36+00:00
    Reply

    Answer:

    5.444\times 10^{-4}

    Explanation:

    The momentum of the neutron before and after the decay  is the same since there’s no external force.

    P_{sys}=const\\\\P=mv\\\\K=0.5mv^2

    #The neutron is initially at rest, so after the decay:

    P_A+P_B=0\\\\P_A=-P_B

    #After decay, the proton has +ve direction  with a velocity v_Awhile the electron moves in a negative direction with a velocity v_B

    Therefore:

    P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B

    Let the energy released during the decay be Q:

    Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}

    Hence,Kp/Ktot is 5.444×10^(-4)

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