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

-5.44x10^(-4)

Answer 1

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 $v_A$while 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 5.444x10^(-4)