1.

The decay constant for the radioactive isotope ""^(57)Co" is "3 xx 10^(-8) s^(-1). The number of disintegrations taking place in a milligram of pure ""^(57)Co per second is :

Answer»

`10^(16)`
`3 xx 10^(11)`
`3 xx 10^(6)`
`3 xx 10^(7)`

Solution :Number of DISINTEGRATION per second is CALLED activity.
Acitivity `=lambdaN`
`=3 xx 10^(-8) xx (6.022 xx 10^(23) xx 10^(-3))/(57)`
`=3.16 x 10^(11)`


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