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The ratio of activities of two radio nuclides A and B in a mixture at initially was found to be 5:1. After two hrs, the ratio of activities become 1: 1. If the \mathrm{t_{1 / 2}} of radio nuclide A is 25 min, then \mathrm{t_{1 / 2}} of the radio nuclide B is (in minutes) : (approx)

Option: 1

50 \mathrm{~min}


Option: 2

30 \mathrm{~min}


Option: 3

40 \mathrm{~min}


Option: 4

25 \mathrm{~min}


Answers (1)

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\mathrm{\frac{A_{0}(A)}{A_{0}(B)}=\frac{5}{1} ; \quad \frac{A_{x}}{A_{y}}=1}

\mathrm{\lambda_{B}-\lambda_{A}=\frac{1}{t} \ln \left[\frac{A_{0}(A)}{A_{0}(B)} \times \frac{A_{x}}{A_{y}}\right] }
\mathrm{0.693\left(\frac{1}{\left(t_{1 / 2}\right)_{B}}-\frac{1}{25}\right)=-\frac{1}{120} \ln 5 }
\mathrm{\frac{25-\left(t_{1 / 2}\right)_{B}}{25\left(t_{1 / 2}\right)_{B}}=-\frac{1.16}{60} \Rightarrow t_{1 / 2}=50 \mathrm{~min} }

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Divya Prakash Singh

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