Question

Flourine 23 has a half-life of 2.2 seconds. how much of a 400 g sample would be left after 11 seconds ?

Answer 1

12.5 grams of Flourine 23 would be left after 11 seconds

Answer 2

Given: Flourine 23 has a half life of 2.2 seconds. Amount of Flourine 23 present at the beginning is 400 gm.

To find: The amount of Flourine 23 left after a duration of 11 seconds.

Solution:

The number of half-lives that 400 gm of Flourine 23 has to go through can be found by the following way,

`(T)/(T_h_a_l_f_-_l_i_f_e) = (11 seconds)/(2.2 seconds) = 5`

So the 400 gm of Flourine 23 will undergo 5 half lives and with each half-life the amount gets divided by 2 or reduces by half.

400 gm -> 200 gm -> 100 gm -> 50 gm -> 25 gm -> 12.5 gm

∴ 12.5 gm of Flourine 23 will be left after 11 seconds.