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

Question

4352 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-13T06:54:33

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

The decay equation is

$m = m_{0} e^{- k t}$ $m = m_0e^(-kt)$ ,

where $m_{0}$ $m_0$ is mass at t = 0.

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

m_0=20 and

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

At t = 1590+50=1640,

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