1.

If 3 xx 10^(-9) kg of radioactive ""_(79)Au^(200) has an activity 58.9 Ci, then half life period of Au^(200) is:

Answer»

`2.88 xx 10^(2)SEC`
`2.88 xx 10^(3)sec`
`2.88 sec`
`2.88 xx 10^(2)sec`

Solution :`N=3 xx 10^(-9) xx ((6.02 xx 10^(23))/(200))`
`=9.03 xx 10^(15)"atoms"`
`R=58.9Ci=58.9 xx 3.7 xx 10^(10)dps`
`=21.8 xx 10^(12)dps`
Now `R=LAMBDA N and T=(0.693)/(lambda)`
`lambda=R/N=(2.18 xx 10^(12))/(9.03 xx 10^(15))=2.41 x 10^(-4) sec^(-1)`
and `T=(0.693)/(lambda)=(0.693)/(2.41 xx 10^(-4))=2.88 xx 10^(3)"sec"`.


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