Home
Class 12
PHYSICS
Activity of a radioactive substance is A...

Activity of a radioactive substance is `A_(1)` at time `t_(1)` and `A_(2)` at time `t_(2)(t_(2) gt t_(1))` , then the ratio of f`(A_(2))/(A_(1))` is:

A

`(t_(2))/(t_(1))`

B

`e^(-lamda(t_(1)+t_(2)))`

C

`e((t_(1)-t_(2))/(lamda))`

D

`e^(lamda(t_(1)-t_(2)))`

Text Solution

Verified by Experts

The correct Answer is:
D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • NUCLEAR PHYSICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE LEVEL-II (LECTURE SHEET (ADVANCED)) Straight Objective Type Questions )|9 Videos
  • NUCLEAR PHYSICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE LEVEL-II (LECTURE SHEET (ADVANCED)) More than one correct answer Type Questions )|4 Videos
  • NUCLEAR PHYSICS

    AAKASH SERIES|Exercise PRACTICE SHEET Matrix Matching Type Questions|2 Videos
  • MOVING CHARGES AND MAGNETISM

    AAKASH SERIES|Exercise EXERCISE-III|49 Videos
  • NUCLEI

    AAKASH SERIES|Exercise Practice Exercise|40 Videos

Similar Questions

Explore conceptually related problems

The activity of a radioactive substance is R_(1) at time t_(1) and R_(2) at time t_(2)(gt t_(1)) . Its decay cosntant is lambda . Then .

The activity of a sample of radioactive material is R_(1) at time t_(1)andR_(2)"at time"t_(2)(t_(2)gtt_(1)) . Its mean life is T. Then,

The activity of a sample of a radioactive meterial is A_1 , at time t_1 , and A_2 at time t_2(t_2 gt t_1) . If its mean life T, then

The radioactivity of a sample is R_(1) at a time T_(1) and R_(2) at time T_(2) . If the half-life of the specimen is T, the number of atoms that have disintegrated in the time (T_(2) -T_(1)) is proporational to

Half-life of a radioactive substance is T. At time t_1 activity of a radioactive substance is R_1 and at time t_2 it is R_2 . Find the number of nuclei decayed in this interval of time.

A radioactive sample S_(1) having the activity A_(1) has twice the number of nucleic as another sample S_(2) of activity A_(2) . If A_(2)=2A_(1) , then the ratio of half-life of S_(1) to the half-life of S_(2) is

Let A_(1) = int_(0)^(x)(int_(0)^(u)f(t)dt) dt and A_(2) = int_(0)^(x)f(u).(x-u) then (A_(1))/(A_(2)) is equal to :

The half-life of a radioactive sample is T . If the activities of the sample at time t_(1) and t_(2) (t_(1) lt t_(2)) and R_(1) and R_(2) respectively, then the number of atoms disintergrated in time t_(2)-t_(1) is proportional to

In a radioactive material the activity at time t_(1) is R_(1) and at a later time t_(2) , it is R_(2) . If the decay constant of the material is lambda , then

Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half- lives T_(1/2)(A): T_(1/2) (B): T_(1/2)(C ) are in the ratio :