Home
Class 12
PHYSICS
Sometimes a radioactive nucleus decays ...

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive . An example is
`.^(38)"Sulphur"underset(=2.48h)overset("half -life")(to) .^(38)Cl underset(=0.62 h)overset("half-life")(to) .^(38)Ar(" stable")`
Assume that we start with `1000 .^(38)S` nuclei at time t=0. the number of `.^(38)Cl` of count zero at t=0 and will again be zero at `t= infty.`At what value of t, would the number of counts be a maximum?

Text Solution

Verified by Experts

The given decay sequence is `.^(38)("Sulper") overset ("half-life") underset(=2.48) to .^(38)Cl overset ("half-life") underset(=0.62h) to .(38)Ar ("stable")`
At any time t, suppose `.^(38)S` has `N_(1) (t)` active nuclei and `.^(38)Cl` have `N_(2)(t)` active nuclei.
`:. (dN_(1))/(dt)=-lambda_(1)N_(1)`=rate of formation of `.^(38)Cl`
and `(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)`= net rate of decay of `.^(38)Cl=-lambda_(2)N_(2)+lambda_(1)N_(0)e^(-lambda_(1)t)`
Multiplying both sides by `e^(lambda_(2)t)dt` and rearranging, `e^(lambda_(2)t)dN_(2)+lambda_(2)N_(2)e^(lambda_(2)t)d t=lambda_(1)N_(0)e^((lambda_(2)-lambda_(1)t)dt`
Integrating both sides, we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^((lambda_(2)-lambda_(1)t)+C.......(i)`
Where C is constant of integration.
At `t=0, N_(2)=0 :. C=-(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))`
Putting in (i), we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^((lambda_(2)-lambda_(1)t))-1)]`
`N_(2)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^(-lambda_(1))-e^(-lambda_(2)t)]`
for maximum count , `N_(2)=max, (dN_(2))/(dt)=0`
`(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(1)t) (-lambda_(1))=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(2)t) (-lambda_(2))` or `(lambda_(1))/(lambda_(2))=(e^(-lambda_(2)t))/(e^(-lambda_(2)t))=e^((lambda_(1)-lambda_(2))t)`
or `log_(e) ((lambda_(1))/(lambda_(2)))=(lambda_(1)-lambda_(2))tlog_(e) e` or `t=(log_(e)(lambda_(1)//lambda_(2)))/((lambda_(1)-lambda_(2)))=(log_(e) T_(2)//T_(1))/(0.693(1/(T_(1))-1/(T_(2))))`
`t=(2.303log_(10) (0.62//2.48)xxT_(1)T_(2))/(0.693 (T_(2)-T_(1)))=(2.303(0-0.602)xx2.48xx0.62)/(0.693(0.62-2.48))=(2.303xx0.602xx2.48xx0.62)/(0.693xx1.86)=1.65sec`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ATOMS AND NUCLEI

    PRADEEP|Exercise Higher order Thinking skills 5|1 Videos
  • ATOMS AND NUCLEI

    PRADEEP|Exercise Higher order Thinking skills 6|1 Videos
  • ATOMS AND NUCLEI

    PRADEEP|Exercise Short question 9|1 Videos
  • COMMUNICATION SYSTEMS

    PRADEEP|Exercise MODEL TEST PAPER-2|9 Videos

Similar Questions

Explore conceptually related problems

A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn disintegrates to some other stable nucleus C. Plot of a graph showing the variation of number of atoms of nucleus B vesus time is : (Assume that at t = 0, there are no B atoms in the sample)

A radioactive substance has half-life T_(1/2) and number of radioactive nuclides present initially was N_(0) . What is the number of radioactive nuclides at time t ?

Knowledge Check

  • A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

    A
    t = T log (1.3)
    B
    `t=T/(log(1.3))`
    C
    `t=T/2 (log 2 )/(log 1.3)`
    D
    `t=T(log 1.3)/(log 2)`
  • A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

    A
    `t = T/(log (1.3))`
    B
    `t = (T/2) (log 2)/(log 1.3)`
    C
    `t = T (log 1.3)/(log 2)`
    D
    `t = T log (1.3)`
  • A radioactive nucleus A with a half life T, decays into nucleus B. At t=0, there is no nucleus B. At somewhat t, the ratio of the number of B to that of A is 0.3 . Then, t is given by

    A
    `t=Tlog(1.3)`
    B
    `t=T/(log (1.3))`
    C
    `t=T/2 (log2)/(log 1.3)`
    D
    `t=T (log1.3)/(log 2)`
  • Similar Questions

    Explore conceptually related problems

    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 :

    The half-life of a sample of a radioactive substance is 1 hour. If 8 xx 10^(10) atoms are present at t = 0, then the number of atoms decayed in the duration t = 2 hours to t = 4 hours will be,

    The half-life of a sample of a radioactive substance is 1 hour. If 8 xx 10^10 atoms are present at t = 0 , then the number of atoms decayed in the duration t = 2 hour to t = 4 hour will be

    In a successive radioactive disintegration underset(N_(1))(A)overset(lamda_(1))tounderset(N_(2))(B)overset(lamda_(2))toC …. No. of nuclides after time t. in which the parent has a longer but not much longer half-life than the daughter. Which of the following expressions is correct?

    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