1.

The decay constant of Ra^(226) " is " 1.37 xx 10^(-1)) s^(-1). A sample of Ra^(226) having an activity of 1.5 millicurie will contain.... atoms

Answer»

`4.1 XX 10^(18)`
`3.7 xx 10^(17)`
`2.05 xx 10^(15)`
`4.7 xx 10^(10)`

SOLUTION :1 milli curie `= 3.7 xx 10^(7) dps`
1.5 milli curie `= 5.55 xx 10^(7) dps`
`(5.55 xx 10^(7))/(N_(0)) = lamda = 1.37 xx 10^(-11)`


Discussion

No Comment Found

Related InterviewSolutions