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 = 4.0…

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asked Jul 28, 2020 134k views

1 Answer

Chris Rogers · Answer 1 · 2020-08-02T06:20:30+0000

Answer: The energy released for the decay of 3 grams of 230-Thorium is

Step-by-step 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

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

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

(Conversion factor:
)

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

$E=\Delta mc^2$

E=(2.324* 10^(-30)kg)* (3* 10^8m/s)^2

E=2.0916* 10^(-13)J

The energy released for 230 grams of decay of thorium is
2.0916* 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* 10^(-13)J

Then, 3 grams of Th will release energy of =
(2.0916* 10^(-13))/(230)* 3=2.728* 10^(-15)J

Hence, the energy released for the decay of 3 grams of 230-Thorium is