Radioactive Decay: Assignment | Modern Physics PDF Download

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 log₁₀ e = 0.4343)

N = N₀e^-λ/t   where λ = 0.693/3.8 = 0.18
N₀/20 = N₀e^-018t  or log₁₀ 20 = 0.18 x t x log₁₀ 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 10³ or A₀ = 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 N₀ be the initial number of nuclei. So, after 2 seconds amount remaining = N₀e^-2λ
∴ Amount dissociated, n = N₀(1 - e^-2λ)    ......(i)
Again after 2 seconds amount remaining = N₀e^-2λ e^-2λ
∴ Amount dissociated, 0.75n = N₀e^-2λ - N₀e^-2λ e^-2λ
⇒ 0.75 = N₀e^-2λ (1 - e^-2λ)   .......(ii)
Solving equations (i) and (ii)

or, 2λ = 2 ln 2 - ln 3 or, 2λ = 2 x 0.6931-1.0986 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 U²³⁸ to Pb²⁰⁶ nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U²³⁸ Take the half-life of U²³⁸ to be 4.5 x 109 year.

= 15 x 10⁹(0.6021 - 0.4771) = 15 x 10⁹ x 0.1250 = 1.875 x 10⁹ 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 N₀ nuclei of the element.
(a) Calculate the number N of nuclei of A at time
(b) If α = 2N₀λ, calculate the number of nuclei of after one half-life 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 N₀ 
(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 - λN₀)  or N = 1/λ [α - (α - (α - N₀λ)e^-λt]   ........(i) 

(b) (i) If α = 2λN0, t = half life = In(2)/λ

(ii) when t → ∞ and α = 2λN₀ 

Q.6. A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 year. Find the time, in year, after which one-fourth of the material remains.

∵ N/N₀ = e^-λt
There is a simultaneous emission of two particles 

Q.7. A small quantity of containing Na²⁴ 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 1cm³ 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 10¹⁰ disintegration per second)

For radioactive decay, R = R₀e^-λt or (-λt) = In R/R₀ 

Half life (T) of the radio nuclide Na²⁴ = 15 hour
Volume of blood = 1 cm³ taken as a sample
After 5 hour, R = 296 dpm in the blood sample.


or R₀ = 296 x 1.26 = 373dpm
or R₀ = 373/60 dps
Activity of one micro curie 10^-6 curie = 3.7 x 10⁴ dps
If activity is (373/60), volume of blood 1 cm³
If activity is dps 3.7 x 10⁴ dps volume 

∴ Volume of blood = 5951.7 cm³ 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 ²³⁸U decays to Pb through a process for which the half-life is 4.5 x 10⁹ year. Find the ratio of number of nuclei of Pb to ²³⁸U after a time of 1.5 x 10⁹ year. Given (2)^1/3 =  1.26.

Radio of number of nuclei of Pb to U²³⁸.
U²³⁸ decays and Pb is formed through the radioactive process.
N = N₀(1/2)ⁿ  where = N₀ 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 10⁹ years and 1.02 x 10⁹ 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 ₉₂U²³⁸ nuclei originally formed is N, the number present now is N₂₃₈ = Ne^-t/T = Ne^-t/6.562 
where t is elapsed time in units of 10⁹ year and T is life time of U. Since the number of ₉₂U²³⁵ nuclei originally formed. The number now present is N₂₃₅ = Ne^-t/1.02 
The present abundance of ₉₂U²³⁵ is


That is, the elapsed time is t = 6.0x10⁹ 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 N₀ = 10²⁰.
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 N₀ = 10
(i) Rate equations:
At an instant t ,  
Let number of nuclei of Y = N_Y and number of nuclei of = N₂ 
∴ 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 = N₀e^-λxt or N_x =(10)²⁰e^{-(0.1)(16.48)}  
or N_X = 1.92 x 10¹⁹ 
Since λ_XN_X = λ_YN_Y according to equation (iv),

or N_Y = 5.76x10¹⁹ 
Again N_Z = 10²⁰ - (1.92 x 10¹⁹) - (5.67 x 10¹⁹)
or N_Z = 10¹⁹ [10 - 1.92 - 5.76]
N_Z = 2.32 x 10¹⁹ 
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 10¹⁹ 
Population of Z = N_Z = 2.32 x 10¹⁹