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radioactive elements help (urgent)
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Amongst the radioactive elements that were incorporated when the Earth was formed were:
uranium 238U with a half-life of 4.5 � 109 y
uranium 235U with a half-life of 7.1 � 108y
thorium 230Th with a half-life of 8.0 � 104 y
iron 60Fe with a half-life of 1.5 � 106 y
manganese 53Mn with a half-life of 3.6 � 106 y
thorium 232Th with a half-life of 13.9 � 109 y
Given that the Earth is believed to have been formed 4600 Ma ago, which of the above isotopes could still be contributing to the radioactive heating of its interior? Could you please give a brief explanation of your reasoning. Thanks.
uranium 238U with a half-life of 4.5 � 109 y
uranium 235U with a half-life of 7.1 � 108y
thorium 230Th with a half-life of 8.0 � 104 y
iron 60Fe with a half-life of 1.5 � 106 y
manganese 53Mn with a half-life of 3.6 � 106 y
thorium 232Th with a half-life of 13.9 � 109 y
Given that the Earth is believed to have been formed 4600 Ma ago, which of the above isotopes could still be contributing to the radioactive heating of its interior? Could you please give a brief explanation of your reasoning. Thanks.
Answers
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Firstly I need to clarify a few points.
I assume that 4.5 x 109 y actually means 4.5 x 10^9 y
I assume 4600 Ma actually means 4600 million years
In that case 4600 Ma = 4.6 x 10^9 y
Now they are in the same units, you can make your comparison.
Firstly I need to clarify a few points.
I assume that 4.5 x 109 y actually means 4.5 x 10^9 y
I assume 4600 Ma actually means 4600 million years
In that case 4600 Ma = 4.6 x 10^9 y
Now they are in the same units, you can make your comparison.
Start by calculating the ratio of what is left
that ratio for what has decayed is is 0.5 raised to the power time/half-life
The time being the age of the Earth
so it's the reciprocal of that
So for the first one that's 0.5 to the power (4.6x10^9/4.5 x10^9) or 0.49 take the reciprocal and 51% of the original U238 remains.
Clearly that's enough to contribute.
Now you do the rest
So
that ratio for what has decayed is is 0.5 raised to the power time/half-life
The time being the age of the Earth
so it's the reciprocal of that
So for the first one that's 0.5 to the power (4.6x10^9/4.5 x10^9) or 0.49 take the reciprocal and 51% of the original U238 remains.
Clearly that's enough to contribute.
Now you do the rest
So