The activity of radionuclide `(""^(100)X)` is 6.023 curie. If the disintegration constatnt is a)`3.7xx10""^(4)sec""^(-1)`,the mass of radionuclide is b)`10""^(-6)`g c)`10""^(-15)`g d)`10""^(-3)`g

Options

A. `10^(-14) g`
B. `10^(-6)` g
C. `10^(-15)` g
D. `10^(-3)` g

Answer

C

Explanation

According to formula `lambda N_(t)=A`
where `lambda` is disintegration constant , `N_(t)` is amount of X left after time t and A is activity in curie. Hence `3.7xx10^(4)xxN_(t)=6.023xx10^(10)` {1 curie `=3.7xx10^(10)` disintegrations per sec}
`therefore N_(t)=(6.023xx3.7xx10^(10))/(3.7xx10^(4))=6.023xx10^(6)`
As the atomic weight of X=100
`6.023xx10^(23)` atoms are in =100g
`therefore 6.023xx10^(6)` atoms will be in
`=(100xx6.023xx10^(6))/(6.023xx10^(23))=100xx10^(-17)g`
`=10^(-15) g`
Hence mass of X will be `10^(-15)` g