Question

The half-life of thorium-227 is 18.72 days. How many days are required for three-fourths given amount to decay?

Answer 1

In `37.44` days.

Explanation:

As half life of Thorium-227 is `18.72` days,

in `18.72` days, half the Thorium-227 is decayed

of remaining half, further half of it i.e. one-fourth of it is decayed in another `18.72` day

Thus three-fourths of given amount is decayed in `18.72xx2=37.44` days.

Mathematically speaking if initial quantity is `Q_0` and half-life is `t_(0.5)` units of time (it could be days, months or years), the quantity left after `t` units of time `Q_t` is given by

`Q_t=Q_0xx2^(-t/t_(0.5))`

Hence `-t/t_(0.5)=log_2(Q_t/Q_0)`

or `t=-t_(0.5)log_2(Q_t/Q_0)`

Here as after `t` days three-fourth is decayed, one-fourth is left i.e. `Q_t=Q_0/4` i.e. `Q_t/Q_0=1/4`

and `t=-18.72xxlog_2(1/4)=-18.72xx(-2)=37.44` days.