Question

Radioactive decay is a first`-` order process. Radioactive carbon in wood sample decays with a half`-` life of 5770 years. What is the rate constant `(` in years `)` for the decay ? What fraction would remains after 11540 years ?

Answer 1

`a.` Given, `t_(1//2)=5770 year`
`K=(0.693)/(t_(1//2))=(0.693)/(5770)`
`=1.2xx10^(-4)year^(-1)`
`b.` `T=nxxt_(1//2)`
`n=(T)/(t_(1//2))=(11540)/(5770)=2`
Now, `(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(2)=(1)/(4)`