Two radioactive nuclei `P` and `Q`, in a given sample decay into a stable nucleus `R`. At time `t = 0`, number of `P` species are `4 N_0` and that of `Q` are `N_0`. Half-life of `P` (for conversation to `R`) is `1mm` whereas that of `Q` is `2 min`. Initially there are no nuclei of `R` present in the sample. When number of nuclei of `P` and `Q` are equal, the number of nuclei of `R` present in the sample would be :

We have two radioactive nuclei A and B . Both convert into a stable nucleus C . Nucleus A converts into C after emitting two two alpha -particles and three beta -particles. Nucleus B converts into C after emitting one. alpha -particle anf five beta -particles. A time t = 0 , nuclei of A are 4N_(0) and that of B are N_(0) . Half-life of A (into the conversion of C ) is 1min and that of B is 2min . Initially number of nuclei of C are zero. What are number of nuclei of C , when number of nuclei of A and B are equal ?

If N_(0) is the number of radioactive nuclei initially present , then the number of nuclei remaining undecayed at the end of nth half life is

Radioactive nuclei P and Q disintegrate into R with half lives 1 month and 2 months respectively. At time t=0 , number of nuclei of each P and Q is x. Time at which rate of disintegration of P and Q are equal , number of nuclei of R is

If p,q and r are prime numbers such that r=q+2 and q=p+2 , then the number of triples of the form (p,q,r) is

There are n number of radioactive nuclei in a sample that undergoes beta decay. If from the sample, n' number of beta -particels are emitted every 2s , then half-life of nuclei is .

A radioactive sample undergoes decay as per the following gragp. At time t=0 , the number of undecayed nuclei is N_(0) . Calculate the number of nuclei left after 1 h . .

Show that the decay rate R of a sample of radionuclide is related to the number of radioactive nuclei N at the same instant by the expression R=lambdaN .

A radioactive element A of decay constant lamda_(A) decays into another radioactive element B of decay constant lamda_(B) . Initially the number of active nuclei of A was N_(0) and B was absent in the sample. The maximum number of active nuclei of B is found at t=2. In 2//lamda_(A) . The maximum number of active nuclei of B is