Answer:
[tex]\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}[/tex]
Explanation:
Given that:
The Half-life of [tex]^{235}U[/tex] = [tex]7.13 \times 10^8 \ years[/tex] is less than that of [tex]^{238} U = 4.51 \times 10^9 \ years[/tex]
Although we are not given any value about the present weight of [tex]^{235}U[/tex].
So, consider the present weight in the percentage of [tex]^{235}U[/tex] to be y%
Then, the time elapsed to get the present weight of [tex]^{235}U[/tex] = [tex]t_1[/tex]
Therefore;
[tex]N_1 = N_o e^{-\lambda \ t_1}[/tex]
here;
[tex]N_1[/tex] = Number of radioactive atoms relating to the weight of y of [tex]^{235}U[/tex]
Thus:
[tex]In( \dfrac{N_1}{N_o}) = - \lambda t_1[/tex]
[tex]In( \dfrac{N_o}{N_1}) = \lambda t_1[/tex] --- (1)
However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of [tex]^{235}U[/tex] to be = [tex]t_2[/tex]
Then:
[tex]In( \dfrac{N_o}{N_2}) = \lambda t_2[/tex] ---- (2)
here;
[tex]N_2[/tex] = Number of radioactive atoms of [tex]^{235}U[/tex] relating to 3.0 a/o weight
Now, equating equation (1) and (2) together, we have:
[tex]In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)[/tex]
replacing the half-life of [tex]^{235}U[/tex] = [tex]7.13 \times 10^8 \ years[/tex]
[tex]In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)[/tex] ( since [tex]\lambda = \dfrac{In 2}{t_{1/2}}[/tex] )
∴
[tex]\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}[/tex]
The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o
Thus, The time elapsed is [tex]\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}[/tex]
