The half-life period of `U^(234)` is `2.5 xx 10^(5)` years. In how much is the quantity of the isotope reduce to 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)) 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 it's rate depends only upon the nature of nucleus and does not depend upon external factors like temoperature an pressure. The rate of radioactive disintigration (Activity) is respresented as (-d(N))/(dt)=lambdaN Where lambda =decay consatant, N= number of nuclei at time t, N""_(0) =initial no. of nucleei. The above equation after integration can be represented as lamda=2.303/tlog.(N""_(0))/(N) Half-life period of U""^(237) is 2.5xx10""^(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) =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

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 ?

The half -life period of ._(84)Po^(210) is 140 days. In how many days 1 g of this isotope is reduced to 0.25 g ?

The half -life period of ._(84)Po^(210) is 140 days. In how many days 1 g of this isotope is reduced to 0.25 g ?

The half-life period of a radioactive element is 1.4 xx 10^(10) years. Calculate the time in which the activity of the element is reduced to 90% of its original value.

The half life period of a first order reaction, A to Product is 10 minutes. In how much time is the concentration of A reduced to 10% of its original concentration?