here's the problem:

"a neutron at rest decays into a proton with a decay energy of 0.78MeV. What is the maximum kinetic energy of the proton left behind?"

here's what I've tried:

In this decay, I assumed that although it's not mentioned, there would be something negatively charged produced to conserve charge. I went with beta decay, so that an electron and an antineutrino would be produced. I assumed that the neutrino was massless and ignored it (my lecturer said it could be ignored).

In the neutron's rest-frame, i said that the proton and electron would have equal and opposite momenta p

_{p}and p

_{e}respectively, ie.

[tex]\stackrel{\rightarrow}{p_e} = -\stackrel{\rightarrow}{p_p}[/tex]

ie.

[tex]p_e = p_p = p[/tex] (magnitudes are equal.)

Also, for conservation of energy, i said that neutron energy E

_{n}= E

_{p}+ E

_{e}= (m

_{p}

^{2}+ p

^{2})

^{1/2}+ (m

_{e}

^{2}+ p

^{2})

^{1/2}= m

_{n}since the neutron has no momentum.

(Here I have used the expression E

^{2}= p

^{2}c

^{2}+ m

^{2}c

^{4}in c=1 units).

m

_{n}- (m

_{p}

^{2}+ p

^{2})

^{1/2}= (m

_{e}

^{2}+ p

^{2})

^{1/2}

(m

_{n}- (m

_{p}

^{2}+ p

^{2})

^{1/2})

^{2}= m

_{e}

^{2}+ p

^{2}

m

_{n}

^{2}- 2m

_{n}(m

_{p}

^{2}+ p

^{2})

^{1/2}+ m

_{p}

^{2}+ p

^{2}= m

_{e}

^{2}+ p

^{2}

m

_{n}

^{2}+ m

_{p}

^{2}- m

_{e}

^{2}= 2m

_{n}(m

_{p}

^{2}+ p

^{2})

^{1/2}

hence

[tex] p = \sqrt{(\frac{m_n^2 + m_p^2 - m_e^2}{2m_n})^2 - m_p^2 } [/tex]

Using wikipedia's data:

m

_{p}= 938.272 MeV/c^2

m

_{n}= 939.566 MeV/c^2

m

_{e}= 0.510 MeV/c^2

I get p = 1.188 MeV/c

Again, using E

^{2}= p

^{2}c

^{2}+ m

^{2}c

^{4}I get the proton energy E

_{p}= 938.2727521 MeV and when I subtract the rest energy from this to get the kinetic energy, I am left with 7.52x10

^{-4}MeV.

When I do the same for the electron, I find that E

_{e}is just less than 0.78 MeV.

Is this a reasonable answer? It seems weird to me that the electron should take the vast majority of the energy, especially when I am looking for the maximum PROTON energy?

Thanks.