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If activity of a radioactive substance's $\;A_{1}\;$ at time $\;t_{1}\;$ & $\;A_{2}\;$ at time $\;t_{2}\;$ $\;(> t_{1})\;$ . Its decay constant is $\;\lambda\;$ . Then

$(a)\;A_{1}=A_{2}\;e^{\lambda(t_{1}-t_{2})}\qquad(b)\;A_{2}=A_{1}^2\;e^{\lambda(t_{1}-t_{2})}\qquad(c)\;A_{1}=A_{2}^2\;e^{\lambda(t_{1}-t_{2})}\qquad(d)\;A_{2}=A_{1}\;e^{\lambda(t_{1}-t_{2})}$

1 Answer

Answer : (d) $\;A_{2}=A_{1}\;e^{\lambda(t_{1}-t_{2})}$
Let $\;A_{0}\;$ be initial activity ,Then
$A_{1}=A_{0}e^{-\lambda t_{1})} \quad \; A_{2}=A_{0} e^{-\lambda t_{2}}$
$\large\frac{A_{2}}{A_{1}}=e^{\lambda(t_{1}-t_{2})}$
$A_{2}=A_{1}e^{\lambda (t_{1}-t_{2})}$
answered Feb 28, 2014 by yamini.v
 

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