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3 years ago

A Decay Chain

Physics

1 answer:

Strike441 [17]3 years ago

7 0

Answer : The energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

Explanation :

First we have to calculate the mass defect $(\Delta m)$ .

The balanced reaction is,

$^{232}Th\rightarrow ^{228}Ra+^{4}He$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})$

$\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg$

conversion used : $(1amu=1.66\times 10^{-27}kg)$

Now we have to calculate the energy released.

$Energy=\Delta m\times (c)^2$

$Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=7.974\times 10^{-13}J$

The energy released is $7.974\times 10^{-13}J$

Therefore, the energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

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Paha777 [63]

Answer:

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Explanation:

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$E_i =E_f$

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Answer:

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