Rate Of Radioactive Disintegration

Radioactive disintegration is a first order reaction and it's 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) = lambda N , 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)) Calculate the half-life period of a radioactive element which remains only 1/16 of it's original amount in 4740 years :

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:

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:

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 it's 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) = lambda N , 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^(232) is 2.5 xx 10^(5) years . In how much time will the amount of U^(237) remaining be only 25% of the original amount ?

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)) What is the activity in Ci (curie) of 1.0mole plutonium -239 ? (t_(1//2)=24000 yeasrs)

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)) Half-life period of U^(2.5xx10^(5) years. In how much thime will the amount of U^(237) remaining be only 25% of the original amount ?

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)) Half-life period of U^(2.5xx10^(5) years. In how much thime will the amount of U^(237) remaining be only 25% of the original amount ?

Radioactive disintegration is a first order reaction and it's 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) = lambda N , 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)) What is the activity in Ci (curie) of 1.0 mole of Plutonium - 239 ? ( t_(1//2) = 24,000 years)

Radioactive Disintegration Series (RDS) || Radioactive Disintegration Law (RDL) || Difference b/w Rate Constant and Disintegration Constant