An isotope has a half-life of 10 minutes. after 20 minutes, what percentage of the original nuclei remain?

To solve this, we can use two equations.

t1/2 = ln 2 / λ = 0.693 / λ

where, t1/2 is half-life and λ is the decay constant.

t1/2 = 10 min = 0.693 / λ

Hence, λ = 0.693 / 10 min - (1)

Nt = Nο e∧ (-λt)

Nt = amount of atoms at t = t time

Nο = initial amount of atoms

t = time taken

by rearranging the equation,

Nt/Nο = e∧ (-λt) - (2)

From (1) and (2),

Nt/Nο = e∧ ( - (0.693 / 10 min) x 20 min)

Nt/Nο = 0.2500

Percentage of remaining nuclei = (nuclei at t time / initial nuclei) x 100%

= (Nt/Nο) x 100%

= 0.2500 x 100%

= 25.00%

Hence, Percentage of remaining nuclei is 25.00%