91Ó°ÊÓ

The reaction energy Q is defined as follows if initial particles A and B interact to form end particles C and D:

$\mathit{Q}\mathbf{=}\mathbf{[}{\mathit{M}}_{\mathbf{A}}\mathbf{+}{\mathit{M}}_{\mathbf{B}}\mathbf{-}{\mathit{M}}_{\mathbf{C}}\mathbf{-}{\mathit{M}}_{\mathbf{D}}\mathbf{]}\mathbf{×}\mathbf{931}\mathbf{.}\mathbf{5}\mathbf{}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u}$

Given;

Atomic mass of ${}_{\mathbf{2}}{}^{\mathbf{4}}\mathit{H}\mathit{e}\mathbf{}\mathrm{is}\mathbf{M}\mathbf{\left(}{}_{\mathbf{2}}{}^{\mathbf{4}}\mathbf{He}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{002603}\mathbf{}\mathbf{u}$.

Atomic mass of ${}_{\mathbf{3}}{}^{\mathbf{7}}\mathit{L}\mathit{i}$is $\mathit{M}\mathbf{\left(}{}_{\mathbf{3}}{}^{\mathbf{7}}\mathit{L}\mathit{i}\mathbf{\right)}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{016005}\mathbf{}\mathit{u}$.

Mass of a neutron is ${\mathit{m}}_{\mathbf{n}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{008665}\mathbf{}\mathit{u}$.

(a) Rewriting the equation as;

${}_{\mathbf{2}}{}^{\mathbf{4}}\mathit{H}\mathit{e}\mathbf{+}{}_{\mathbf{3}}{}^{\mathbf{7}}\mathit{L}\mathit{i}\mathbf{→}{}_{\mathbf{Z}}{}^{\mathbf{A}}\mathit{X}\mathbf{+}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathit{n}$

Therefore, from the conservation of the mass number and the atomic no.

Hence, Z and A for the nuclide X is 5 and 10.

Boron is the periodic table element with Z=5; As a result, the nuclide X ${}_{\mathbf{5}}{}^{\mathbf{10}}\mathit{B}$has an atomic mass of $\mathit{M}\mathbf{\left(}{}_{\mathbf{5}}{}^{\mathbf{10}}\mathit{B}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{012937}$

Now, write the reaction with Boron as follows;

${}_{\mathbf{2}}{}^{\mathbf{4}}\mathit{H}\mathit{e}\mathbf{+}{}_{\mathbf{3}}{}^{\mathbf{7}}\mathit{L}\mathit{i}\mathbf{→}{}_{\mathbf{5}}{}^{\mathbf{10}}\mathit{B}\mathbf{+}{}_{\mathbf{0}}{}^{\mathbf{1}}\mathit{n}$

Because the amount of electrons in this equation is equal on both sides, the atomic masses of these elements can be used because the electron masses cancel each other out.

Putting the values;

$$\begin{gathered} \mathit{Q}\mathbf{=}\mathbf{[}\mathbf{4}\mathbf{.}\mathbf{002603}\mathbf{}\mathit{u}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{016005}\mathbf{}\mathit{u}\mathbf{-}\mathbf{10}\mathbf{.}\mathbf{012937}\mathbf{}\mathit{u}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{008665}\mathbf{]}\mathbf{×}\mathbf{931}\mathbf{.}\mathbf{5}\mathbf{}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{[}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{002994}\mathbf{}\mathit{u}\mathbf{]}\mathbf{×}\mathbf{931}\mathbf{.}\mathbf{5}\mathbf{}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u} \\ \mathit{Q}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{789}\mathbf{}\mathit{M}\mathit{e}\mathit{V} \end{gathered}$$

The reaction is absorbed because the reaction energy is negative (endoergic).

Hence, the energy is Q = - 2.789 and it is absorbed.