# The half-life of radium is 1590 year. If 10 grams is present now, how much will be present in 50 years?

Jan 13, 2017

$20 {\left(\frac{1}{2}\right)}^{\frac{1640}{1590}} = 9.78439$ gm, nearly.7.13 gm, nearly

#### Explanation:

The decay equation is

$m = {m}_{0} {e}^{- k t}$,

where ${m}_{0}$ is mass at t = 0.

Given #m_0/2=10 ahen t = 1590,

m_0=20 and

$10 = 20 {\left({e}^{-} k\right)}^{1590}$, giving ${e}^{- k} = {\left(\frac{1}{2}\right)}^{\frac{1}{1590}}$

At t = 1590+50=1640,

$m = 20 {\left({e}^{-} k\right)}^{1640} = 20 {\left(\frac{1}{2}\right)}^{\frac{1640}{1590}} = 9.78439$ gm, nearly.