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.

Step-by-step 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)* c^2`

`E=(0.001407u)* (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.