1.

The radioactivity of a sample is R_1 at a time t_1 and R_2 at time t_2. If half life of sample is T, then no. of atoms that have disintegrated in time (t_2 - t_1) is proportional

Answer»

`R_(1)t_(1)-R_(2)t_(2)`
`(R_(1)-R_(2))^(-1)`
`(R_(1)-R_(2))/(T)`
`(R_(1)-R_(2))T`

Solution :`R_(1)=LAMBDA N_(1), R_(2)=lambda N_(2)`
As `N_(2) lt N_(1)`, so no. of atoms decayed in TIME `(t_(2)-t_(1))` is
`N_(1)-N_(2)=(R_(1))/(lambda)=(R_(1))/(lambda) -R_(2)/lambda =((R_(1)-R_(2))/(lambda))`
`rArr N_(1)-N_(2)=((R_(1)-R_(2))T)/(0.693) ALPHA (R_(1)-R_(2))T`


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