Question

the half-life of a radioisotope 63 hours how much time will elapse if the radioisotope decays to one-fourth of its original mass?

Answer 1

Ans:

Let the original mass be m1

Let the molar mass of the isotope be ‘M’.

No. of isotope initially, N_o = (m1/M) x N_A                               {N_A = Avagadro’s no = 6.023 x 10²³}

Mass after the decay, m2 = m1/4

No. of isotopes after the decay, N_t = (m2/M) x N_A = (m1/4M) x N_A

Half-life, t_½ = 63 hours

λ = decay constant = 0.693/t_½ = 0.693/63 = 0.011 hours^-1

We know that all radioactive decay follows first order kinetics. Accordingly, we have

N_t = N_o . e^(-λt)

where, Nt = no of radioactive isotopes after the decay

            N_o = Initial no of radioactive isotopes

            λ = decay constant

t = time taken for decay =?

(m1/4M) x N_A = {(m1/M) x N_A} e^{(-0.011 x t)}

t = 126 hours

OR

In a first order reaction for original mass to become half, it takes 1 half-life. For original mass to become 1/4^th of its original mass it takes 2 half-life.

So, t = 2t_½ = 2 x 63 = 126 hours