InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The radionuclide `.^(56)Mn` is being produced in a cyclontron at a constant rate `P` by bombarding a manganese target with deutrons. `.^(56)Mn` has a half-life of `2.5 h` and the target contains large numbers of only the stable manganese isotopes `.^(56)Mn`. The reaction that produces `.^(56)Mn` is `.^(56)Mn +d rarr .^(56)Mn +p` After being bombarded for a long time, the activity of `.^(56)Mn` becomes constant, equal to `13.86 xx 10^(10) s^(-1)`. (Use `1 n2=0.693`, Avagardo number `=6 xx 10^(2)`, atomic weight of `.^(56)Mn=56 g mol^(-1)`). After a long time bombardment, number `.^(56)Mn` nuclei present in the target depends upon.A. All (i), (ii) and (iii) are correctB. Only (i) and (ii) are correctC. Only (ii) and (iii) are correctD. Only (i) and (iii) are correct. |
|
Answer» Correct Answer - C As `N=(Pt_(1//2))/(ln 2)`. It is dependent upon P and `t_(1//2)`. In equilibrium, rate of production=rate of decay. Large initial number will only make equilibrium come sonner. Hence, Initial number of `.^(56)Mn` nuclei will make no difference |
|