Radio activity

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A radioactive sample of U²³⁸ decay to Pb through a process for which half life is 4.5 x 10⁹ years. The ratio of number of nuclei of Pb to U²³⁸ after a time of 1.5 x 10⁹ years (given 2^1/3 = 1.26)

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Solution

$Here\,\,\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} = {\left( {\frac{1}{2}} \right)^{1/3}}$
$Where\,\,n = Number\,\,of\,\,half\,\,lives = \frac{1}{3}$
$\Rightarrow \frac{N}{{{N_0}}} = \frac{1}{{1.26}} \Rightarrow \frac{{{N_U}}}{{{N_{Pb}} + {N_U}}} = \frac{1}{{1.26}}$
$\Rightarrow {N_{Pb}} = 0.26\,{N_U} \Rightarrow \frac{{{N_{Pb}}}}{{{N_U}}} = 0.26$

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Radio activity

Remember concepts with our Masterclasses.

A radioactive sample of U238 decay to Pb through a process for which half life is 4.5 x 109 years. The ratio of number of nuclei of Pb to U238 after a time of 1.5 x 109 years (given 21/3 = 1.26)

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A radioactive sample of U²³⁸ decay to Pb through a process for which half life is 4.5 x 10⁹ years. The ratio of number of nuclei of Pb to U²³⁸ after a time of 1.5 x 10⁹ years (given 2^1/3 = 1.26)