Question

Bromine- is radioactive and has a half life of seconds. What percentage of a sample would be left after seconds? Round your answer to significant digits.

Answer 1

Corrected question:

Bromine-88 is radioactive and has a half life of 16.3seconds. What percentage of a sample would be left after 35seconds? Round your answer to 2 significant digits.

Answer:

23% (3 significant digits)

Explanation:

$M_{R} =\frac{M_{O} }{2^{n} }$

where $M_{R}$ is mass remaining

          $M_{O}$ is original mass

          $n=\frac{t}{t1/2}$

       t1/2= 16.3s  ,  t=35s

n =35/16.3 =2.147

$M_{R} =\frac{M_{O} }{2^{2.147} }$

$M_{R}$ = $0.2258M_{O}$

0.2258*100% =22.5%

≈23%