Sol. Since, at t = 0, probabilities of getting α and β particles from the radionuclide are equal, therefore, initial activities of two isotopes are equal. Let it be A 0 .

Activity of first isotope at t = 1620 sec

A 1 = A 0 =

That of second isotope, A 2 = A 0 =

∴ Total activity of radionuclide at t = 1620 sec, A = A 1 + A 2 = A 0

∴ Probability of getting α -particle, P 1 = =

and that of getting β –particle, P 2 = =

Let at t = 0, number of nuclei of two isotopes be N 01 and N 02 respectively.

Initial activity of first isotope, A 1 = N 01 λ 1 = N 01

That of second isotope, A 2 = N 02 λ 2 = N 02

Since, A 1 = A 2 , therefore, =

Or = =

Initially, total number of nuclei, N 0 = N 01 + N 02

∴ N 01 = N 0 and N 02 = N 0

At time t, number of nuclei of first isotope that remain undecayed,

N 1 = N 01 = N 0

That of second isotope, N 2 = A 02 = N 0

∴ Total number of nuclei remaining undecayed at time t,

N = N 1 + N 2 = + N 0 = (x 4 + 4x)

Where x =

But it is equal to N 0 /2.

∴ (x 4 + 4x) = or x 4 + 4x – 2.5 = 0

Hence, x = 0.594 or = 0.594

Thanking log, – log 2 = log 0.594

or – × 0.30103 = (– 1 + 0.7742275)

or t = 1215 seconds