Question

Suppose the half-life of an element is 10 years. How many half-lives will it take before only about 6% of the original sample remains? 3 half-lives 4 half-lives 5 half-lives 6 half-lives

Answer 1

We can solve this problem by using the equation as shown: A=Aoe^-(ln2/t)T where A is the final amount (0.06 gram), Ao is the initial amount (1g *assumed), t is the half-life (10 years ) and lastly T time elapsed (required). Plug in the values in the equation and you should get  40.5 years and this is approximately 4 half-lives. 

Answer 2

Answer is: 4 half-lives will it take before only about 6% of the original sample remains.

Take 100 atoms:

After first half-life: 50% · 100 ÷ 100% = 50.

After second half-life: 0.5 · 50 = 25; numbo of atoms after secend half-life.

After third half-life: 0.5 · 25 = 12.5.

After fourth half-life:  0.5 · 12.5 = 6.25.

6.25 ÷ 100 · 100% = 6.25%.

Half-life is the time required for a quantity (in this example number of radioactive nuclei of lincolnium) to reduce to half its initial value.