Q.1. The half life of radioactive radon is 3.8 day. Find the time at the end of which (1 / 20)^{th} of the radon sample will remain undecayed. (Given log_{10} e = 0.4343)
N = N_{0}e^{λ/t} where λ = 0.693/3.8 = 0.18
N_{0}/20 = N_{0}e^{018t} or log_{10} 20 = 0.18 x t x log_{10} e
or 1.3 = 0.18 x 0.4343 x t
Q.2. A 280 day old radioactive substance shows an activity of 6000 dps, 140 day later its activity becomes 3000 dps. What was its initial activity?
For radioactive disintegration,
or A0 = 24 x 10^{3} or A_{0} = 24000
∴ Initial activity = 24000 dps
Q.3. A radioactive sample emits nβ particles in 2 sec. In next 2 sec it emits 0.75nβ particles. What is the mean life of the sample? Given ln 2 = 0.6931 and ln 3 = 1.0986
Let N_{0} be the initial number of nuclei. So, after 2 seconds amount remaining = N_{0}e^{2λ}
∴ Amount dissociated, n = N_{0}(1  e^{2λ}) ......(i)
Again after 2 seconds amount remaining = N_{0}e^{2λ} e^{2}^{λ}
∴ Amount dissociated, 0.75n = N_{0}e^{}^{2λ}  N_{0}e^{2λ} e^{2}^{λ}
⇒ 0.75 = N_{0}e^{2λ} (1  e^{2λ}) .......(ii)
Solving equations (i) and (ii)
or, 2λ = 2 ln 2  ln 3 or, 2λ = 2 x 0.69311.0986 or, λ = 0.1438sec^{1}
∴ Mean life = 1/λ = 6.954 second or mean life = 7 second
Q.4. In an ore containing uranium, the ratio of U^{238} to Pb^{206} nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U^{238} Take the halflife of U^{238} to be 4.5 x 109 year.
= 15 x 10^{9}(0.6021  0.4771) = 15 x 10^{9} x 0.1250 = 1.875 x 10^{9} year
Q.5. Nuclei of a radioactive element A are being produced at a constant ratea α. The element has a decay constant λ. At time t = 0 , there are N_{0} nuclei of the element.
(a) Calculate the number N of nuclei of A at time
(b) If α = 2N_{0}λ, calculate the number of nuclei of after one halflife of A, and also the limiting value of N as t → ∞
Nuclei of a radioactive element A are being produced at a constant ratea .
Decay constant of element = λ ,
At t = 0 , nuclei of element present N_{0}
(a) Number N of nuclei of A at time t :
Net rate of formation of nuclei of element A = dN/dt
or α  λN = e^{λt}(α  λN_{0}) or N = 1/λ [α  (α  (α  N_{0}λ)e^{λt}] ........(i)(b) (i) If α = 2λN0, t = half life = In(2)/λ
(ii) when t → ∞ and α = 2λN_{0}
Q.6. A radioactive material decays by simultaneous emission of two particles with respective halflives 1620 and 810 year. Find the time, in year, after which onefourth of the material remains.
∵ N/N_{0} = e^{λt}
There is a simultaneous emission of two particles
Q.7. A small quantity of containing Na^{24} radio nuclide (half life = 15 hour)of activity 1.0 micro curie is injected into the blood of a person A sample of the blood of volume 1cm^{3} taken after 5 hour shows an activity of 296 disintegration per minute. Determine the total volume blood in the body of the person. Assume that radioactive solution mixes uniformly in the blood of the person.
(1curie = 3.7 x 10^{10} disintegration per second)
For radioactive decay, R = R_{0}e^{λt} or (λt) = In R/R_{0}
Half life (T) of the radio nuclide Na^{24 }= 15 hour
Volume of blood = 1 cm^{3} taken as a sample
After 5 hour, R = 296 dpm in the blood sample.
or R_{0} = 296 x 1.26 = 373dpm
or R_{0} = 373/60 dps
Activity of one micro curie 10^{6} curie = 3.7 x 10^{4} dps
If activity is (373/60), volume of blood 1 cm^{3}
If activity is dps 3.7 x 10^{4} dps volume
∴ Volume of blood = 5951.7 cm^{3} or Volume of blood = 5.95litre
∴ Total volume of blood in the body of person is 5.95 litre.
Q.8. A radioactive sample of ^{238}U decays to Pb through a process for which the halflife is 4.5 x 10^{9} year. Find the ratio of number of nuclei of Pb to ^{238}U after a time of 1.5 x 10^{9} year. Given (2)^{1/3 }= 1.26.
Radio of number of nuclei of Pb to U^{238}.
U^{238} decays and Pb is formed through the radioactive process.
N = N_{0}(1/2)^{n} where = N_{0} Initial number of radioactive material
N = Number of radioactive material left, n = Number of half lives
Q.9. In the mixture of isotopes normally found on the earth at the present time, has an abundance of 99.3% and has an abundance of 0.7%. The measured lifetimes of these isotopes are 6.52 x 10^{9} years and 1.02 x 10^{9} years, respectively. Assuming that they were equally abundant when the earth was formed, the estimate age of the earth, in years.
If the number of _{92}U^{238} nuclei originally formed is N, the number present now is N_{238} = Ne^{t/T} = Ne^{t/6.562 }
where t is elapsed time in units of 10^{9} year and T is life time of U. Since the number of _{92}U^{235} nuclei originally formed. The number now present is N_{235} = Ne^{t/1.02}
The present abundance of _{92}U^{235} is
That is, the elapsed time is t = 6.0x10^{9} yr.
Q.10. A radioactive nucleus X decays to a nucleus Y with a decay constant λ_{X} = 0.1s^{1}. Y further decays to a stable nucleus Z with a decay constant λ_{Y} = (1/30)s^{1} Initially there are only X s nuclei and their number is N_{0} = 10^{20}.
Set, up the rate equations for the populations of X , Y and. Z. The Populations of Y nucleus as a function ‘of time is given by
Find the time at which N_{Y} is maximum and determine the populations of X and Z at that instant.
Radioactive nucleus X decays to a nucleus Y
Y further decays to a stable nucleus Z
Initially there are only X nuclei and there number is N_{0} = 10
(i) Rate equations:
At an instant t ,
Let number of nuclei of Y = N_{Y} and number of nuclei of = N_{2}
∴ Rate equations of populations of X , Y and Z are(ii) Time t at which N_{Y} is maximum:
or t = 15ln(3) or t = 16.48 sec
(iii) Populations of X and Z when N_{Y} is maximum:
N_{X} = N_{0}e^{λxt} or N_{x} =(10)^{20}e^{(0.1)(16.48) }
or N_{X} = 1.92 x 10^{19}
Since λ_{X}N_{X} = λ_{Y}N_{Y} according to equation (iv),
or N_{Y} = 5.76x10^{19}
Again N_{Z} = 10^{20}  (1.92 x 10^{19})  (5.67 x 10^{19})
or N_{Z} = 10^{19} [10  1.92  5.76]
N_{Z} = 2.32 x 10^{19}
Hence (i) Rate equations for populations X, Y, Z are given in equations (i), (ii) and (iii)
(ii) Time at which N_{Y} is maximum = 16.48sec
(iii) Population of X = N_{X} = 1.92 x 10^{19}
Population of Z = N_{Z} = 2.32 x 10^{19}
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