If decay constant of a radioactive sample is 0.05 per year, then find the time (in year) for which sample will decay by 75%.

Question

If decay constant of a radioactive sample is 0.05 per year, then find the time (in year) for which sample will decay by 75%. (A) 28 (B) 56 (C) 14 (D) 7

Tardigrade Admin · Accepted Answer

Here, λ = 0.05 per year N=N0-(3N0/4)=N0/4, t=? loge (N/N0)=λ t t=(loge (N0/N)/λ)=(loge 4/0.05)=27.72 year ≈28 year

Q. If decay constant of a radioactive sample is 0.05 per year, then find the time (in year) for which sample will decay by 75%.

28

56

14

7

Here, $λ$ = 0.05 per year
$N = N_{0} - \frac{3 N _{0}}{4} = N_{0} /4, t = ?$
log $_{e} \frac{N}{N _{0}} = λ t$
$t = \frac{l o g _{e} \frac{N _{0}}{N}}{λ} = \frac{l o g _{e} 4}{0.05} = 27.72$ year $\approx 28$ year

Q. If decay constant of a radioactive sample is 0.05 per year, then find the time (in year) for which sample will decay by 75%.

28

56

14

7

Here, λ = 0.05 per year N=N0​−43N0​​=N0​/4,t=? loge​N0​N​=λt t=λloge​NN0​​​=0.05loge​4​=27.72 year ≈28 year

Here, $λ$ = 0.05 per year
$N = N_{0} - \frac{3 N _{0}}{4} = N_{0} /4, t = ?$
log $_{e} \frac{N}{N _{0}} = λ t$
$t = \frac{l o g _{e} \frac{N _{0}}{N}}{λ} = \frac{l o g _{e} 4}{0.05} = 27.72$ year $\approx 28$ year