1.

The half life of a substance is 20 minutes. E The time interval between 33% decay and 67% decay.

Answer»

SOLUTION :For 33% DECAY, `N/N_0=1/2^(n_1) RARR 67/100=1/2^(n_1)`….(1)
For 67% decay, `N/N_0=1/2^(n_2) rArr 33/100 = 1/2^(n_2)` ….(2)
`n_2=t_2/T`
`(2)/(1)=33/67 =1/(2^(n_2-n_1))`
`rArr 1/(2)^1 =1/(2(t_2-t_1)/T) rArr (t_2-t_1)/T=1`
`t_2-t_1`= T=20 minutes.


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