Question

How much energy is released by the decay of 3 grams of 230Th 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 for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$

Explanation:

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

The equation for the alpha decay of thorium nucleus follows:

$_{90}^{230}\textrm{Th}\rightarrow _{88}^{226}\textrm{Ra}+_2^{4}\textrm{He}$

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

$\Delta m=(m_{Ra}+m_{He})-(m_{Th})$

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

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

To calculate the energy released, we use Einstein equation, which is:

$E=\Delta mc^2$

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

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

The energy released for 230 grams of decay of thorium is $2.0916\times 10^{-13}J$

We need to calculate the energy released for the decay of 3 grams of thorium. By applying unitary method, we get:

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

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

Hence, the energy released for the decay of 3 grams of 230-Thorium is $2.728\times 10^{-15}J$