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**Hint:**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.

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