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A radioactive material of Lay life T was produced in a nuclear reactor at different instants , the quantity produced second time was twice of that produced first time . If now their present activities are $\;A_{1}\;$ and $\;A_{2}\;$ respectively . Then their age difference equals :

$(a)\;\large\frac{T}{ln 2}\;|ln \large\frac{2A_{1}}{A_{2}}|\qquad(b)\;\large\frac{T}{ln 2}\;|ln \large\frac{A_{2}}{2 A_{1}}|\qquad(c)\;T |ln \large\frac{A_{1}}{A_{2}}|\qquad(d)\;T |ln \large\frac{A_{2}}{2 A_{1}}|$

1 Answer

Answer : (b) $\;\large\frac{T}{ln^2}\;|ln \large\frac{A_{2}}{2 A_{1}}|$
Explanation :
Since $\;A_{1}=\lambda N_{0} e^{-\lambda t_{1}}$
or $\;t_{1}=\large\frac{1}{\lambda} ln(\large\frac{\lambda N_{0}}{A_{1}})$
$A_{2}=(\lambda)\;(2 N_{0}) e^{-\lambda t_{2}}$
$t_{2}=\large\frac{1}{\lambda} ln (\large\frac{2 \lambda N_{0}}{A_{2}})$
$t_{1}-t_{2}=\large\frac{1}{\lambda} ln(\large\frac{A_{2}}{2 A_{1}})$
$=\large\frac{T}{ln 2 (\large\frac{A_{2}}{2 A_{1}})}$
answered Feb 27, 2014 by yamini.v

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