0.0625g remains from 1g of a radioactive element after 20 years of radioactive decay. Determine the rate constant and half-life `(t_(1//2))` of the reaction. How much of the element did remain after 10 years from the start?

Explanation

Radioactive disintegration process follows a first order kinetics . For a first order kinetics , the integrated form of the rate equation is `lamda=(2.303)/(t)log.(N_(0))/(N)`
where `lamda` = disintegration constant (or rate constant), `N_(0) and N` are the number of radio active atoms at t = 0 and at time t ,respectively.
According to the problem, `(N_(0))/(N)=(1)/(0.0625)` , when t = 20y
Substituting the values for `(N_(0))/(N)` and t into the above
equation gives `lamda=(2.303)/(20)log.(1)/(0.0625)=0.1386y^(-1)`
So , the rate constant of the reaction = `0.1386y^(-1)`
and half-life `=(0.693)/(lamda)=(0.693)/(0.1386)=5y`
If Wg of the element remains after 10y, then `(N_(0))/(N)=(1)/(W)`
Therefore , `lamda=0.1386=(2.303)/(10)log.(1)/(W) " or, " W=0.25`
Thus , 0.25g of the element will remain after 10 y.