Error in calculating binding energy of Beryllium 8 [closed]

I don't know where I am going wrong with regards to calculating the binding energy of $^8$Be. This is what I am doing: I look up online and find that the mass excess of $^8$Be is $5305 \mu$u. Thus, the atomic mass is ($8 + 5305\mu$)u. The nuclear mass is therefore (because of $4$ electrons): $8 + (5305 - 4 \times 548.6)\times 10^{-6}$ = $8.003$u. We have $4$ protons and $4$ neutrons, so their rest mass is: $4(1.0073 + 1.0087) = 8.064$ u. The mass deficit is therefore: $8.064 - 8.003 = 0.0609$ u. This corresponds to a binding energy of $0.0609 \times 931.49 = 56.72$ MeV. However, if I compare this with literature I get a binding energy of $56.50$ MeV. Can someone please tell me where I am going wrong?

Citations are below: http://dbserv.pnpi.spb.ru/elbib/tablisot/toi98/www/astro/table2.pdf (I can't link cite for mass excess of 8Be as that is through my uni login)

closed as off-topic by sammy gerbil, user36790, Wolpertinger, John Rennie, NeuneckAug 19 '16 at 10:10

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Try $m_p = 1.0072765$ u, $m_n = 1.0086649$ u and $m_e = 0.000554558$ u with $1$ u = $1.66053904 \times 10 ^{-27}$ kg or $1$ u $= 931.490954$ MeV/c$^2$ and $m_\text{Be8} = 8.0053051$ u.
• Thanks a lot bro. I get $56.499$ MeV now which is almost exact. I find it weird how in the textbook they give it to 5sf so I should only use all masses and constants to 5sf too, but it is not that accurate then. Also, I can see now in nuclear physics you have to be crazy accurate too! – Vaibhav Sharma Aug 18 '16 at 11:41