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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 reactio, indicate the apporipriate target nucleus. If the half life of `._(27)^(60)Co` is 7 year evaluate the decay constant in `s^(-1)`.

Answer» `._(28)^(60)Ni + ._(0)^(1)n rarr ._(27)^(60) Co + ._(+1)^(1)P`
`._(27)^(59) Co + ._(o)n^(1) rarr ._(27)^(60) Co + gamma`
Thus, target nucleaus is `._(28)^(60)Ni` for `(n,p)` and `._(27)^(59)Co` for `(n, gamma)` reaction.
`lambda = (0.693)/(t_(1//2))`
`= (0.693)/(7xx365xx24xx60xx60)`
`= 3.14xx10^(-9) sec^(-1)`


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