Question

In the reaction $\mathbf{p}\mathbf{+}{}^{\mathbf{19}}\mathbf{F}\mathbf{\to }\mathbf{\alpha }\mathbf{+}{}^{\mathbf{16}}\mathbf{O}$, the masses are m(p) = 1.007825 u, m(a) = 4.002603 u, m(F) = 18.998405 u, m(O) = 15.994915 u. Calculate the Q of the reaction from these data.

Short answer

The Q value of the reaction is 8.12 MeV.

Step-by-step solution

Identification of given data

The mass of Fluorine nuclei is $\mathrm{m}\left(\mathrm{F}\right)=18.998405\mathrm{u}$

The mass of Oxygen nuclei is $\mathrm{m}\left(\mathrm{O}\right)=15.994915\mathrm{u}$

The mass of proton is $\mathrm{m}\left(\mathrm{p}\right)=1.007825\mathrm{u}$

The mass of alpha particle is $\mathrm{m}\left(\mathrm{\alpha }\right)=4.002603\mathrm{u}$

The Q value of the reaction is the energy released due to the mass defect from reactant and products in reaction.

Determination of Q value of the reaction

The Q value of the reaction is given as:

$\mathrm{Q}=\left[\left(\mathrm{m}\left(\mathrm{p}\right)+\mathrm{m}\left(\mathrm{F}\right)\right)-\left(\mathrm{m}\left(\mathrm{\alpha }\right)+\mathrm{m}\left(\mathrm{O}\right)\right)\right]{\mathrm{c}}^2$

Here, c is the speed of light and its value in terms of mass and energy is 931.5 MeV/u

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{E}=\left[\left(1.007825\mathrm{u}+18.998405\mathrm{u}\right)-\left(4.002603\mathrm{u}+15.994915\mathrm{u}\right)\right]\left(\frac{931.5\mathrm{MeV}}{\mathrm{u}}\right) \\ \mathrm{E}=8.12\mathrm{MeV} \end{gathered}$$

Therefore, the Q value of the reaction is 8.12 MeV.