Question

What percent of a present amount of radioactive radium (226Ra) will remain after 1200 years? The half life of 226Ra is 1599 years.

Answer 1

59.44%

Explanation:

`M_(R) = (M_(O) )/(2^(n) )`

`n=(t)/(t(1)/(2) )`

Where
`M_(R)` = mass remaining

`M_(O)` =original mass.

t= time

`t(1)/(2)` =half life

`t(1)/(2)` =1599, t= 1200,
`M_(O)` =226,
`M_(R)` =?

`n=(1200)/(1599)`

n= 0.750469

`M_(R) = (226)/(2^( 0.750469) )`

`M_(R)` =134.3367amu

percentage of mass remaining =
`(134.3367amu)/(226amu)` *100%

= 59.44%