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Ede4ka [16]
3 years ago
5

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)
Physics
1 answer:
Katyanochek1 [597]3 years ago
7 0

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

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)\times c^2

E=(0.003785u)\times (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.

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Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

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Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

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m is the mass of the moon

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U2 = -GMm/(r2+r1)

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∆U = -GMm{1/(r2+r1)-1/r1}

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G = 6.67×10^-11m³/kgs²

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r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0
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