I have the following reaction: 105Be+ 21H→ 115B+ 11H
And I know that I have to use the formula: E=Δm⋅c2=Δm⋅931,5MeVu.
So I just need Δm which is equal to: Δm=mb−ma where mb represents the mass "before the reaction" and ma the mass "after the reaction" so we have:
mb=m(105Be)+m(21H)
ma=m(115B)+m(11H)
The book which contains this problem contains the following table: http://i.imgur.com/esoGDVf.png but from this table, I only know m(11H) and m(21H) i.e.
mb=m(105Be)+2.01410u
ma=m(115B)+1.00783u
How do I calculate m(105Be) and m(115B) ?
P.S. I don't know if the tag is correct. The chapter in the book where I found this exercise is called "Basics of nuclear physics".
Answer
How do I calculate m(104Be) and m(115B)?
The masses and various other properties of isotopes are available freely at Wolfram Alpha. They are,
- m(104Be)=10.013533818u
- m(115B)=11.009305406u
where u denotes unified atomic mass units. Notice you are already given the mass number in the superscript of the isotope. As John Rennie noted, the reaction should probably be with 94Be,
94Be+21H→105B+n
in which case the mass is 9.012182201u.
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