Home
Class 12
PHYSICS
Consider a radioactive disintegration ac...

Consider a radioactive disintegration according to the equation `ArarrBrarrC`. Decay constant of A and B is same and equal to `lambda`. Number of nuclei of A, B and C are `N_0` , 0, 0 respectively at `t=0`. Find
(a) number of nuclei of B as function of time t.
(b) time t at which the activity of B is maximum and the value of maximum activity of B.

Text Solution

Verified by Experts

The correct Answer is:
(a) `N_(B) = lamdaN_(0) ("te"^(-lamdat))` (b) `t = 1/lamda , R_(max) = (lamdaN_(0))/e`
Promotional Banner

Similar Questions

Explore conceptually related problems

Two radioactive materials A & B have decay constant 3lamda and 2lamda respectively. At t=0 the numbers of nuclei of A and B are 4N_(0) and 2N_(0) respectively then,

Number of nuclei of a radioactive substance are 1000 and 900 at times t=0 and time t=2 s . Then, number of nuclei at time t=4s will be

At time t=0 , number of nuclei of a radioactive substance are 100. At t=1 s these numbers become 90. Find the number of nuclei at t=2 s .

At time t=0 , number of nuclei of a radioactive substance are 100. At t=1 s these numbers become 90 . Find the number of nuclei at t=2 s .

Nucleus A decays into B with a decay constant lamda_(1) and B further decays into C with a decay constant lamda_(2) . Initially, at t = 0, the number of nuclei of A and B were 3N_(0) and N_(0) respectively. If at t = t_(0) number of nuclei of B becomes constant and equal to 2N_(0) , then

A radio nuclide X is produced at constant rate alpha . At time t=0 , number of nuclei of X are zero. Find (a) the maximum number of nuclei of X. the number of nuclei at time t. Decay constant of X is lambda .

Two radioactive samples of different elements (half-lives t_1 and t_2 respectively) have same number of nuclei at t=0 . The time after which their activities are same is

Nucleus A decays to B with decay constant lambda_(1) and B decays to C with decay constant lambda_(2) . Initially at t=0 number of nuclei of A and B are 2N_(0) and N_(0) respectively. At t=t_(o) , no. of nuclei of B is (3N_(0))/(2) and nuclei of B stop changing. Find t_(0) ?

The decay constant of a radioactive isotope is lambda . If A_1 and A_2 are its activites at time t_1 and t_2 respectively, then the number of nuclei which have decayed the time (t_1-t_2)

Consider radioactive decay of A to B with which further decays either to X or Y , lambda_(1), lambda_(2) and lambda_(3) are decay constant for A to B decay, B to X decay and Bto Y decay respectively. At t=0 , the number of nuclei of A,B,X and Y are N_(0), N_(0) zero and zero respectively. N_(1),N_(2),N_(3) and N_(4) are the number of nuclei of A,B,X and Y at any instant t . The number of nuclei of B will first increase and then after a maximum value, it decreases for