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Home » Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example isAssume that we start with 1000 38S nuclei at time . The number of is of count zero at and will again be zero at . At what value of , would the number of counts be a maximum?
Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
Sulphur^{38}\frac{\text { half-life }\longrightarrow}{=2.48 h}Cl^{38}\frac{\text { half-life }\longrightarrow}{=0.62 h}Ar^{38}(Stable)
Assume that we start with 1000 38S nuclei at time t=0. The number of Cl^{38} is of count zero at t=0 and will again be zero at t=\infty. At what value of t, would the number of counts be a maximum?
CBSE | Class 12 | NCERT Exemplar | Nuclei | Physics
Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
Sulphur^{38}\frac{\text { half-life }\longrightarrow}{=2.48 h}Cl^{38}\frac{\text { half-life }\longrightarrow}{=0.62 h}Ar^{38}(Stable)
Assume that we start with 1000 38S nuclei at time t=0. The number of Cl^{38} is of count zero at t=0 and will again be zero at t=\infty. At what value of t, would the number of counts be a maximum?

Let the disintegration constants for S^{38} and Cl^{38} be \lambda_ 1 and \lambda_ 2 respectively.

\mathrm dN_1 / \mathrm{dt}=-\lambda N_{1}

\mathrm dN_2 / \mathrm{dt}= rate of decay of Cl^{38}+ rate of formation of Gl^{38}

\mathrm dN_2 / \mathrm{dt}=-\lambda_ 2 \mathrm N_2+\lambda_ 1 N_1

\mathrm{e}^{\lambda 2 t} \mathrm{~d} \mathrm N_2+\lambda_ 2 \mathrm N_2 \mathrm{e}^{\lambda 2 \mathrm{t}} \mathrm{dt}=\lambda_ 1 \mathrm{N_oe}^{(\lambda_ 2-\lambda_ 1)} \mathrm{dt}

Integrating the above equation we get,

t=1.65 \mathrm{~h}

i

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