Half-life of thorium is `1.5xx10^10` y. How much time is needed for 20% of thorium to disintegrate ?

Explanation

Let initial mass of thorium =`N_0`
If in time t 20% of the thorium is disintegrated then,
the amount of thorium that disintegrates
`=N_0xx20/100=0.2N_0`
Amount of thorium left,
`N=N_0-0.2 N_0 = 0.8 N_0`
Now, `N=N_0e^(-lambdat)` or , `e^(lambdat ) = N_0/N=N_0/(0.8N_0)`=1.25
`therefore lambdat = log_e(1.25) ` = 0.223
or , `0.693/T ` t=0.223 [`because` T(half-life ) =`0.693/lambda` ]
or, `t=T/0.693xx0.223`
`=(1.5xx10^10xx223)/693=0.48xx10^10` y (approx. )
Alternative method :
`N=0.8N_0`
Also `N=N_0/2^(t//T)`
or , `0.8=1/2^(t//T)` or , `2^(t//T)` = 5/4
or , `t//T=log_2 5//4 =(log_10 5//4)/(log_10 2)=0.0969/0.3010`=0.322
or , t=0.322 T = `0.322 xx 1.5 xx 10^10` y
`=0.48xx10^10` y (approx.)