Question

Question #480ff

Answer 1

The amount remaining will be 2 %.

The number of half-lives is `n = t/t_½`

For each half-life, you divide the total amount of the isotope by 2, so

Amount remaining = `"original amount"/2^n` or

`A = A_0/2^n`

`n = t/t_½ = (11.8"h")/(2"h")` = 5.90 half-lives (1 significant figure + 2 guard digits)

`A = A_0/2^n = (100"%")/2^5.90= (100"%")/59.7` = "1.67" % = 2 % (1 significant figure)

You can't work this out exactly without a calculator, because the time is not an integral multiple of the half-life.

If the time had been 12 h, that would have been 6 half-lives.

Then you could have written

100 % × `1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2 = (100"%")/64` = "1.56" % = 2 % (1 significant figure)

Hope this helps.

http://socratic.org/chemistry/nuclear-chemistry/nuclear-half-life-calculations/half-life-calculations-part-3-easy-way