Question

How much energy is released by the decay of 3 grams of 20Th in the following reaction 230 Th - 226Ra + 'He (230 Th = 229.9837 g/mol, 226Ra - 225.9771 g/mol, "He = 4.008 g/mol) (10 pts.)

Answer 1

Answer : The energy released by the decay of 3 grams of 'Th' is $2.728\times 10^{-15}J$

Explanation :

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

The balanced reaction is,

$^{230}Th\rightarrow ^{226}Ra+^{4}He$

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

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

$\Delta m=(225.9771+4.008)-(229.9837)=1.4\times 10^{-3}amu=2.324\times 10^{-30}kg$

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

Now we have to calculate the energy released.

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

$Energy=(2.324\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=2.0916\times 10^{-13}J$

The energy released is $2.0916\times 10^{-13}J$

Now we have to calculate the energy released by the decay of 3 grams of 'Th'.

As, 230 grams of Th release energy = $2.0916\times 10^{-13}J$

So, 3 grams of Th release energy = $\frac{3}{230}\times 2.0916\times 10^{-13}J=2.728\times 10^{-15}J$

Therefore, the energy released by the decay of 3 grams of 'Th' is $2.728\times 10^{-15}J$