Question

Determine Q for the reaction${}_3{}^{7}Li+{}_1{}^{1}H\to {}_4{}^{7}Be+{}_0{}^{1}n$

Short answer

Energy released$Q=-1.64MeV$

Step-by-step solution

Given data and concept

The reaction between lithium and hydrogen:

$L{i}_3^7+H_1^1\to B{e}_4^7+n_0^1$

• $m_{L_3^7}=6.941u$- the mass of lithium
• $m_{H_1^1}=1.0079u$- the mass of hydrogen
• $m_{B{e}_4^7}=7.01692u$- the mass of beryllium
• $m_{n_0^2}=1.00866u$- unknown

So, we had to determine the released heat Q for the reaction.

We will be applying an equation that determines energy as:

$Q=\Delta m{c}^2$

Where:$\Delta m$ - mass defect C - speed of light

Where the mass defect $\Delta m$is given by: $\Delta m=m_{L_3^7}+m_{H_1^1}-m_{B{e}_4^7}-m_{n_0^2}{c}^2$.

Put known values for finding Q

For calculating the Q for the reaction:

$\begin{array}{rcl}Q& =& (6.941u+1.0079u-7.01692u-1.00866u)c^2\\ & =& -0.00176\times 931.5MeV\\ & =& -1.64MeV\\ & & \end{array}$

Therefore, the amount of energy released is$Q=-1.64MeV$