Question

Calculate the $Q$ value in the following decay using the atomic masses given in the table.
${}^{25}\text{Al}{\to }^{25}\text{Mg}+e^{+}+\nu$

Mass of ${}^{25}\text{Al}$	Mass of ${}^{25}\text{Mg}$
$24.990432u$	$24.985839u$

$[\text{Take}{m}_{electron}=0.511\frac{\text{MeV}}{c^2};c^2=931\frac{\text{MeV}}{u}]$

• A. $14.816\text{MeV}$
• B. $12.408\text{MeV}$
• C. $11.204\text{MeV}$
• D. $3.254\text{MeV}$

Answer 1

The correct option is D $3.254\text{MeV}$

The given decay is a ${\beta }^{+}$decay.

The $Q$ value of ${\beta }^{+}-decay$ is given by,

$Q=\left[m{(}^{25}\text{Al}\right)-m{(}^{25}\text{Mg})-2m_{electron}]c^2$

$\Rightarrow Q=[24.99032u-24.985839u-2\times 0.511\frac{\text{MeV}}{c^2}]c^2$

$\Rightarrow Q=\left(0.004593u\right)\left(931\frac{\text{MeV}}{u}\right)-1.022\text{MeV}$

$\Rightarrow Q=4.276\text{MeV}-1.022\text{MeV}$

$\Rightarrow Q=3.254\text{MeV}$

Hence, option $\left(D\right)$ is correct.

Why this Question? ${\beta }^{+}-\text{decay}:$ ${}_Z{X}^A{\to }_{Z-1}{Y}^A{+}_{+1}{e}^0+\nu +Q$ In ${\beta }^{+}$ decay, inside a nucleus a proton is converted into a neutron, positron and neutrino. $p\to n{+}_{-1}{e}^0+v$