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A parent nucleus \mathrm{x} is decaying into daughter nucleus \mathrm{y} which in furn decays to \mathrm{z}. The half lives of \mathrm{x} and \mathrm{y} are \mathrm{40000} yeas and \mathrm{20} years respectively. In a certain sample, it is found that the number of \mathrm{y} nuclei hardly changes with time. If the number of \mathrm{x} nuclei in the sample is \mathrm{4 \times 10^{20}}, the number of \mathrm{y} nuclei present in it is-

Option: 1

\mathrm{2\times10^{7}}


Option: 2

\mathrm{2\times10^{17}}


Option: 3

\mathrm{4\times10^{23}}


Option: 4

\mathrm{4\times10^{20}}


Answers (1)

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\mathrm{\begin{aligned} & \lambda_x N_x=\lambda_y N_y \\ & N_y=\frac{\lambda_x}{\lambda_y} \cdot N_x \end{aligned}}
\mathrm{=\frac{\left(t_{1 / 2}\right) y}{\left(t_{1 / 2}\right) x} \cdot N_x}
\mathrm{\begin{aligned} & N_y=\frac{20}{40000} \times 4 \times 10^{20} \\ & N_y=2 \times 10^{17} \quad \text { } \end{aligned}}

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Rakesh

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