[" Question 2) Half life period "],[" of "U^(234)" is "2.5times10^(5)" years.In "],[" how many years it will "],[" remain "25%" of original "],[" amount? "]

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?

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 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 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) =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 radium is 1600 years . In how many years 1 g radium will be reduced to 0.125 g ?