Question

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

Answer 1

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%