Question

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The atomic mass of 40 K 40 K is 39.963998 u and that of 40 C a 40 C a is 39.962591 u)

Answer 1

Answer: The energy released in the given nuclear reaction is 1.3106 MeV.

Explanation:

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$    (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.