What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca is 41.958618 u)

Question

asked May 22, 2020 26.6k views

1 Answer

Nemanja Banda · Answer 1 · 2020-05-29T09:04:12+0000

Answer: The energy released in the given nuclear reaction is 3.526 MeV.

Step-by-step explanation:

For the given nuclear reaction:

$_(19)^(42)\textrm{K}\rightarrow _(20)^(42)\textrm{Ca}+_(-1)^(0)\textrm{e}$

We are given:

Mass of
$_(19)^(42)\textrm{K}$ = 41.962403 u

Mass of
$_(20)^(42)\textrm{Ca}$ = 41.958618 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:

$\Delta m=(41.962403-41.958618)=0.003785u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.003785u)* c^2$

E=(0.003785u)* (931.5MeV) (Conversion factor:
)

E=3.526MeV

Hence, the energy released in the given nuclear reaction is 3.526 MeV.