The half life of radioactive Radon is `3.8 days` . The time at the end of which `(1)/(20) th` of the radon sample will remain undecayed is `(given log e = 0.4343 ) `
A
`3.8 days`
B
`16.5 days`
C
`33 days`
D
` 76 days`
Text Solution
Verified by Experts
The correct Answer is:
B
(b) t_(1//2) = 3.8 `days `:. Lambda = (0.693)/(t_(1//2)) = (0.693)/(3.8) = 0.182` If the initial number of aton is `a = A_(0)` then after time `t` the number of aloms is `20 = A` . We have to find `t` `t = (2.303)/(lambda) log (A_(0))/(A) = (2.303)/(0.182) log (a)/(a//20) = 16.46 days`
