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It is observed that only 0.39% of the or...

It is observed that only 0.39% of the original radioactive sample remains undecayed after eight hours.Hence:

A

the half-life of that substance is 1 hour

B

the mean life of the substance is `(1)/(ln 2)` hour

C

decay constant of the substance is (ln2) `"hour"^-1`

D

if the number of radioactive nuclei of this substance at a given instant is `10^8` then the number left after 30 min would be `sqrt2 xx 10^(9)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(or) `((0.693)/(tau_(1//2))) xx 8 = 2 .303 log ((100)/(0.39)) , ((0.693)/(tau_(1//2))) xx 8 = 2.303 log ((100 xx 100)/(39))`
`tau_(1//2) = (0.693 xx 8)/(2.303 xx 2.4089) -= 1 , tau_(1//2) = 1 , lambda = 0.693 = I_(n) 2 , tau = (1)/(lambda) = (1)/(0.693) = (1)/(I_(n) 2)`
`(2.303 log 2) xx (1)/(2) = 2 .303 log ((10^(8))/(N_(t))) , (t)/(2) log 2 = log ((10^(8))/(N_(t))) log (2)^(t//2) = log "" (10^8)/(N_t) , (10^8)/(N_t) = sqrt2 implies N_t = (1)/(sqrt2) 10^8`
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