Question

The half-life of \(^{210}_{84}po\) is 138 d. Find the time required for 75% of the initial number of radio active nuclei of \(^{210}_{84}po\) to disintegrate.

Answer 1

Data: T = 138 d, = \(\cfrac{N}{N_0}\) = \(\cfrac14\) (since only 25% of the initial number of nuclei remains undisintegrated)

In one half-life (t = T), \(\cfrac{N}{N_0}\) = \(\cfrac{1}{2}\).

In two halflives (t = 2T),

\(\cfrac{N}{N_0}\) = \(\cfrac{1}{4}\)

∴ The time for 75% of \(^{210}_{84}po\) nuclei to disintegrate is 2T = 2 × 138 = 276 d.