homework and exercises - How to calculate the energy freed in the reaction: $^{10}_5Be +space ^2_1H rightarrow space^{11}_5B + space ^1_1H$?

I have the following reaction: $^{10}_5\mathrm{Be} +\space ^2_1\mathrm{H} \rightarrow \space^{11}_5\mathrm{B} + \space ^1_1\mathrm{H}$

And I know that I have to use the formula: $E = \Delta m\cdot c^2 = \Delta m \cdot \frac{931,5MeV}{u}$.

So I just need $\Delta m$ which is equal to: $\Delta m = m_b - m_a$ where $m_b$ represents the mass "before the reaction" and $m_a$ the mass "after the reaction" so we have:
$m_b = m(^{10}_5\mathrm{Be}) + m(^2_1\mathrm{H})$
$m_a = m(^{11}_5\mathrm{B}) + m(^1_1\mathrm{H})$

The book which contains this problem contains the following table: http://i.imgur.com/esoGDVf.png but from this table, I only know $ m(^1_1\mathrm{H})$ and $m(^2_1\mathrm{H})$ i.e.
$m_b = m(^{10}_5\mathrm{Be}) + 2.01410u$
$m_a = m(^{11}_5\mathrm{B}) + 1.00783u$

How do I calculate $m(^{10}_5\mathrm{Be})$ and $m(^{11}_5\mathrm{B})$ ?

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 $m\left(^{10}_4\mathrm{Be}\right)$ and $m\left(^{11}_5 \mathrm{B}\right)$?

The masses and various other properties of isotopes are available freely at Wolfram Alpha. They are,

• $m\left(^{10}_4\mathrm{Be}\right)=10.013533818u$


• $m\left(^{11}_5 \mathrm{B}\right)=11.009305406u$

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 $^{9}_4\mathrm{Be}$,

$$^{9}_4\mathrm{Be} + ^{2}_1 \mathrm{H} \to ^{10}_{5}\mathrm{B} + n$$

in which case the mass is $9.012182201u$.