Deduce the expression, `N=N_(0e)^(-lambdat)`, for the law of radioactive decay.

Deduce the expression, N=N_(0)^(-lambdat) for the law of radioactive decay. (b) Is the nucleus formed in the decay of the nucleus ._(11)^(12)Na , an isotope or isobar?

Radioactive Decay Law

Expand the following N=N_(0)e^(-lambdat)

Radioactive decay follows first-order kinetic. The mean life and half-life of nuclear decay process are tau = 1// lambda and t_(1//2) = 0.693//lambda . Therefore are a number of radioactive elements in nature, their abundance is directly proportional to half life. The amount remaining after n half lives of radioactive elements can be calculated using the relation: N = N_(0) ((1)/(2))^(n) The rate of radioactive decay is

Stability Curve and Laws of Radioactive Decay

The mathematical equation of law of radioactive decay is

Radioactive disintergation always follow I order kinetics and is independent of all external factors and is represented by the relation N = N_(0) e^(-lambda t) where lambda is decay constant and N atoms are left at time t . The radioactive nature of element is expressed in terms of average life numerically equal to decay constant (1//lambda) however all the radioactive do not lose their radioactive nature in thier average life. The radioactive emission involves alpha, beta particles as well as gamma- rays.The penertrating power order is alpha lt beta lt gamma . The emissions can perntrate even thick steel walls but are however unable to penttrate Pb blocks. The S unit fo rate of decay is dps . The percentage of atoms decayed in average life of a radioactive element is: