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* 10^(-15)J`

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:
`1amu=1.66* 10^(-27)kg` )

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
`2.728* 10^(-15)J`