Barium-141 has a half-life of 18 minutes.The half-life of Krypton-92 is 3.0 s .In the fission reaction of a mass of Uranium - 235 , equual numbers of barium and krypton nuclei are produced .Estimate the time tken after the fission of the sample of uranium for the ratio
`("number of Barium" - 141 "nuclei")/("number of" "Krypton"-92 "nuclei")`. to be approximately equal to 8.

Two radioactive nuclei X and Y initially contain an equal number of atoms.Their half life is 1 hour and 2 hours respectively.Calculate ratio of their rates of disintegration after two hours.

Radioactive element A are being produced at a constant rate alpha . The element has a decay constant lambda . At time t=0, there are N_0 nuclei of the element. If alpha=2N_0lambda , calculate the number of nuclei of A after one half life of A and also limiting value of N as t rarr infty .

The isotope Phosphorus-33 undergoes beta - decay to form Sulphur-33,which is stable.The half - life of Phosphorus-33 is 24.8 days. A pure sample of Phosphorus - 33 has an initial activity of 3.7 xx 10^6 Bq. Calculate the initial number of Phosphorus - 33 nuclei in the sample,

Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Calculate the half-life period of a radioactive element which remains only 1//16 of its original amount in 4740 years: a) 1185 years b) 2370 years c) 52.5 years d) none of these

Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =initial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Half-life period of U is 2.5xx10^(5) years. In how much time will the amount of U^(237) remaining be only 25% of the original amount ? a) 2.5xx10^(5) year b) 1.25xx10^(5) years c) 5xx10^(5) years d) none of these

The isotope Phosphorus-33 (_(15)P^(33)) undergoes beta - decay to form "Sulphur"-33(_(16)S^(33)) ,which is stable.The half - life of Phosphorus-33 is 24.8 days. A pure sample of Phosphorus - 33 has an initial activity of 3.7 xx 10^6 Bq.Calculate the number of Phophorus - 33 nuclei remaining in the sample after 30 days.