Four thousand active nuclei of a radioactive material are present at textt = 0. After 60 minutes 500 active nuclei are left in the sample. The decay constant of the sample is

Question

Four thousand active nuclei of a radioactive material are present at textt = 0. After 60 minutes 500 active nuclei are left in the sample. The decay constant of the sample is (A) (ln (2)/60) per minute (B) (ln (2)/20) per minute (C) 20 ln (2) per minute (D) 60 ln (2) per minute

Tardigrade Admin · Accepted Answer

60 minutes = 3 half lives ∴ T(1/2)=20 min ∴ λ =(ln 2/t1 / 2)=(ln 2/20)per min

Q. Four thousand active nuclei of a radioactive material are present at $t = 0.$ After $60$ minutes $500$ active nuclei are left in the sample. The decay constant of the sample is

$\frac{l n ( 2 )}{60}$ per minute

$\frac{l n ( 2 )}{20}$ per minute

$20 l n (2)$ per minute

$60 l n (2)$ per minute

60 minutes = 3 half lives

$∴ T_{\frac{1}{2}} = 20 min$

$∴ λ = \frac{l n 2}{t _{1/2}} = \frac{l n 2}{20} p er min$

Q. Four thousand active nuclei of a radioactive material are present at t=0. After 60 minutes 500 active nuclei are left in the sample. The decay constant of the sample is

60ln(2)​ per minute

20ln(2)​ per minute

20ln(2) per minute

60ln(2) per minute