A radioactive nucleus X with a half-life T , decays into a nucleus Y. At t = 0 , there is no nucleus Y. After some time ' t ', the ratio of the number of Y to that of X is 0.4 . Then, '

Question

A radioactive nucleus X with a half-life T , decays into a nucleus Y. At t = 0 , there is no nucleus Y. After some time ' t ', the ratio of the number of Y to that of X is 0.4 . Then, ' t ' is

TestHub · Accepted Answer

Let the initial number of nuclei of X be $N_{0}$ .

At time $t$ , the number of nuclei of X is $N_{X}$ and the number of nuclei of Y is $N_{Y} = N_{0} - N_{X}$ .

Given $\frac{N _{Y}}{N _{X}} = 0.4$

$\frac{N _{0} - N _{X}}{N _{X}} = 0.4$

$\frac{N _{0}}{N _{X}} - 1 = 0.4$

$\frac{N _{0}}{N _{X}} = 1.4$

$N_{X} = \frac{N _{0}}{1.4}$

Using the radioactive decay law, $N_{X} = N_{0} (\frac{1}{2})^{\frac{t}{T}}$

$\frac{N _{0}}{1.4} = N_{0} (\frac{1}{2})^{\frac{t}{T}}$

$1.4 = 2^{\frac{t}{T}}$

Taking logarithm on both sides:

$lo g (1.4) = \frac{t}{T} lo g (2)$

$t = \frac{T lo g ( 1.4 )}{lo g ( 2 )}$

Answer: $\frac{T lo g ( 1.4 )}{lo g ( 2 )}$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion