Question

8. The majority of naturally occurring rhenium is 1877sRe, which is radioactive and has a half-life of 7x1010 years. In how many years will 5% of the earth's 187sRe decompose?

Answer 1

The half-life period of ¹⁸⁷₇₅Re = 7 x 10¹⁰ years

The initial amount of ¹⁸⁷₇₅Re(N_o) = 100 %

determine the amount of ¹⁸⁷₇₅Re remains after 5% decomposition.

The amount of ¹⁸⁷₇₅Re after time "t" (N_t) = 100 % -5% = 95 %

Determine the decay constant(k) by using the half-life period as follows:

k = 0.693 / t_1/2

Substitute 7 x 10¹⁰ years for t_1/2 and solve for decay constant(k) as follows:

k = 0.693 / 7 x 10¹⁰ years

k = 9.9 x 10^-12 year^-1

Now, the nuclear decomposition reactions are first order reactions.

Thus, the first-order decay equation is as follows:

ln[N_t / N₀] = -kt

Rearrange the formula for "t" as follows:

t = -[1/k] x ln[N_t/N_o]

Substitute the known values and solve for "t"as follows:

t = -[1/9.9 x 10^-12 year^-1] x ln[95%/100%]

t = 5 x 10⁹ years

Thus, In 5 x 10⁹ years, 5% of earths ¹⁸⁷₇₅Re will decompose.