The count rate observed from a radioacti...

The count rate observed from a radioactive source at t second was `N_0` and at 4t second it was `(N_0)/(16)`. The count rate observed at `(11/2)t` second will be

`(N_0)/(128)`

`(N_0)/(64)`

`(N_0)/(32)`

None of these

Text Solution

Verified by Experts

The correct Answer is:

`(N_0)/(16)=N_0(1/2)^n` `:.` `n=4`
So, 3t times is equivalent to four half-lives. Hence, one half-life is equal to `(3t)/(4)`.
The given time `11/2t-t=9/2t` is equivalent to 6 half-lives.
`:.` `N=N_0(1/2)^6=(N_0)/(64)`

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Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 seconds it was 25600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to:

150

400

360

200

The initial count rate for a radioactive source was found to be 1600 count/sec.and t = 8 sec this rate was 100 count/sec. What was the count rate at t = 6 [in counts/sec]-

`4000`

`300`

`200`

`150`

The count rate of a radioactive sample was 1600 count/s at t=0 and 100 counts at t=8 s . Select the correct alternatives

Its count rate was 200 counts at t=6 s

Its count rate was 200 counts at t=4 s

Half life of the sample is 4s

Mean life of the sample is 2.88 s