Question

The time in which the activity of an element reduces to 90% of its original value is:
The half-life period of the element is $1.4 \times {10}^{10} years$.
A. $2.128 \times {10}^{9} years$
B. $1.578 \times {10}^{7} years$
C. $6.954 \times {10}^{7} years$
D. None of these

Answer 1

The radioactive decay constant is defined as the probability of a given unstable nucleus decaying per unit time. It is denoted by $\lambda$. The mathematical formula of decay constant is $\lambda \quad =\quad \cfrac { 0.693 }{ T }$

Complete step by step answer: It is given in the question that activity of the element is reduced to 90% of its original value which means that 10% of the element is used. The reaction here that is taking place is a first order reaction.
Therefore, the half-life (T) for a first order reaction is given as the ratio of 0.693 to the decay constant.
$T = \cfrac { 0.693 }{ \lambda }$
$\implies \lambda = \cfrac { 0.693 }{ T }$
It is given that the half-life of the element is $1.4 \times {10}^{10} years$. Substituting this value in this above equation, we get
$\lambda = \cfrac { 0.693 }{ 1.4 \times { 10 }^{ 10 } }$
Now, the kinetic equation for the first order reaction is given as
$t = \cfrac { 2.303 }{ \lambda } \log { (\cfrac { { N }_{ 0 } }{ N } } )$ -----(1)
Where $\lambda$ is the decay constant, ${N}_{0}$ is the initial concentration at time 0, N is the final concentration at time t and t is the activity time.
Let us assume that the initial concentration was 100. Thus, the final concentration will be 90 and $\lambda = \cfrac { 0.693 }{ 1.4 \times { 10 }^{ 10 } }$. Substituting these values in equation (1), we get
$td = \cfrac { 2.303 }{ \cfrac { 0.693 }{ 1.4 \times { 10 }^{ 10 } } } \log { (\cfrac { 100 }{ 90 } } )$
$\implies t = \cfrac { 2.303 }{ 0.693 } \times 1.4\quad \times { 10 }^{ 10 } \times \log { (\cfrac { 10 }{ 9 } } )$
$\implies t = 2.128 \times { 10 }^{ 9 } years$

Therefore, the activity time is $2.128 \times { 10 }^{ 9 } years$. Hence, the correct answer is option (A).

Note: While calculating the decay constant, do make sure that the base of the log is 10 else you might end up getting the wrong answer. The half-life of a sample is 69.3% of the mean life which is reciprocal of the decay constant. It is applicable for any sample.