Question

(b) A radioactive isotope has a half-life of $10$ years. How long will it take for the activity to reduce to $3.125$%?

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Updated on : 2022-09-05

Solution

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A. The value of binding energy per nucleon gives a measure of the stability of that nucleus greater is the binding energy per nucleon of a nucleus , more stable is the nucleus.

The above graph shown, binding energy per nucleon drawn against mass number A.

The saturation effect of nuclear forces is the property responsible for the approximate constancy of binding energy in range 30<A<17030

It is clear from the curve that binding energy per nucleon of the fused nucle is more than there of the light nuclei taking part nuclear fusion .

Hence energy gets released in the process.

In a nuclear fission , the sum of the masses of the final products is less than the sum of the masses of the reactant component

B. $N=N_{o}e_{−λt}$ But $N=32N_{o} $

$32N_{o} =N_{o}e_{−λt}$

$2_{−5}=e_{−λt}$

5 ln2 = \lambda t$$

but given half life so $λ=10ln2 $

so t = 50 Years

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