The radioactive isotope `._(27)^(60)Co` which has now replaced radium in the treatment of cancer can be made by `a(n,p)` or `(n, gamma)` reaction. For each reaction, indicate the appropriate target nucleus. If the half life of `._(27)^(60)Co` is 7 year evaluate the decay constant in `s^(-1)`.

Answer

`3.142xx10^(-9)s^(-1)`

Explanation

`a.` `i..(n.p)` `._(27)Co^(60)ii. ....(n,gamma)` `._(27)Co^(60)+` `._(1)H^(1)`
`i. ` `._(28)Ni^(60)+` `underset((Targ et n ucl eus i s ._(28)Ni^(60)))(._(0)n^(1)rarr ` `._(27)Co^(60)+ ._(1)H^(1))`
`ii.` `._(27)Co^(59)+` `underset((Targ et n ucl eus is ._(27)Co^(59)))(._(0)n^(1)rarr ` `._(27)Co^(60)+` `._(1)H^(1))`
`b.` `t_(1//2)` of `._(27)Co^(60)=7` years
`K=(0.693)/(t_(1//2))`
`=(0.693)/((7years)xx(365 day year^(-1))(24 hr day^(-1))(60 mi n hr^(-1))(60 mi ns^(-1)))`
`=(0.693)/(7xx3.15xx10^(7)s)`
`=31.42xx10^(-10)s^(-1)`
`=3.142xx10^(-9)s^(-1)`