Home
Class 12
PHYSICS
A parent radioactive nucleus A (decay co...

A parent radioactive nucleus A (decay constant `lamda_(A))` converts into a radio-active nucleus B of decay constant `lamda_(b)`, initially, number of atoms of B is zero At any time `N_(a),N_(b)` are number of atoms of nuclei A and B respectively then maximum velue of `N_(b)`.

A

`(3)/(2)`

B

`(9)/(4)`

C

`(1)/(1)`

D

`(ln3)/(ln2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR PHYSICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE PRACTICE SHEET (ADVANCED) More than one correct answer Type Questions|5 Videos

  • NUCLEAR PHYSICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE PRACTICE SHEET (ADVANCED) Linked Comprehension Type Questions Passage-I :|2 Videos

  • NUCLEAR PHYSICS

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE LEVEL-II (LECTURE SHEET (ADVANCED)) Integer Type Questions )|3 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

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

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

Radioactive material 'A' has decay constant '8 lamda' and material 'B' has decay constant 'lamda'. Initial they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be (1)/( e) ?

Radioactive material 'A' has decay constant '8 lamda' and material 'B' has decay constant 'lamda'. Initial they have same number of nuclei. After what time, the ratio of number of nuclei of material 'A' to that 'B' will be '(1/ e) ?

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,

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) ?

A radioacitve nucleus is being produced at a constant rate alpha per second. Its decay constant is lambda . If N_(0) are the number of nuclei at time t=0 , then maximum number of nuceli possible are .

A radioactive nucleus is being produced at a constant rate alpha per second. Its decay constant is lamda . If N_0 are the number of nuclei at time t = 0 , then maximum number of nuclei possible are.

A radioactive substance "A" having N_(0) active nuclei at t=0 , decays to another radioactive substance "B" with decay constant lambda_(1) . B further decays to a stable substance "C" with decay constant lambda_(2) . (a) Find the number of nuclei of A, B and C time t . (b) What should be the answer of part (a) if lambda_(1) gt gt lambda_(2) and lambda_(1) lt lt lambda_(2)

The plot of the number (N) of decayed atoms versus activity (R) of a radioactive substance is