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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)

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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:
1u=931.5MeV/c^2 )


E=3.526MeV

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

User Nemanja Banda
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