Class 12 Physics Nuclei | Terminologies related to Radioactive decay |

**Terminologies related to Radioactive decay**

- Decay Rate: -
- Decay rate is defined as the number of nuclei decaying per unit time.
- Denoted by ‘R’.
- Mathematically:- R =-(dN/dt)
- Where dN=change in the number of nuclei with time and (-ive) sign shows the number of nuclei decreasing.
- From Exponential decay law N = N
_{0}e^{-λt}, therefore - R=(d/dt)( N
_{0}e^{-λt})

- =- N
_{0}[e^{-λt}(-λ)] - R= N
_{0}λ e^{-λt}; - At t=0 , R
_{0}= N_{0}λ (equation (1)) - Therefore
**R=R**=>decay constant also changes exponentially._{0}e^{-λt}

- Activity of radioactive sample:-
- Activity of radioactive sample is defined as the total decay rate R of a sample of one or more radionuclides.
- Decay rate of whole sample is considered.
- I. unit: - Becquerel(Bq); 1Bq = (1 decay/sec).
- Other unit: - Curie(Ci); 1C i=3.7 x10
^{10}

- Half-life of radioactive sample:-
- Half – life of radioactive sample is defined as the time at which number of nuclei reduces to one-half of their initial values.
- It is denoted by t (
_{1/2}). - Half-life tells how long radioactive nuclei can last.
- Mathematical expression:-
- From radioactive law , N=N
_{0}e^{-λt}(equation(1)) - Initially number of sample = N
_{0}. - Later number of samples will become=(N
_{0}/2). - Therefore from equation(1)- (N
_{0}/2) =N_{0}e^{-λt}_{(1/2)} - =>e
^{-λt}_{(1/2)}= (1/2) - =>- λt
_{(1/2)}= ln(1/2) (taking log on both sides) - =>- λt
_{(1/2)}=ln(1) – ln(2) - =>λt
_{ (1/2)}= ln(2) (using ln (1) =0). - =>
**t**_{(1/2)}= (0.693/λ)

- From radioactive law , N=N
- Mean-life of radioactive sample:-
- Mean life is defined as the average life of a nuclei in the radioactive sample.
- Any nuclei can decay at any interval of time.
- It is denoted by t
_{av}. - Mathematically:-
- Let number of radioactive sample at t=0 =N
_{0}. - From the law of radioactive decay (dN/dt) =-λN
- => dN= -λNdt (these many nuclei will decay in time dt).
- Therefore number of nuclei which decay between t and (t+dt) =λNdt.
- But of the nuclei will decay fast and some will slowly decay.
- Sum of lives of all these nuclei = tλNdt.
- Therefore average life of the sample =t
_{av}= (_{0}^{∞}∫(tλN)dt/N_{0}) - = (λ/N
_{0})_{0}^{∞}∫ (tN) dt. - From exponential law of decay :- N =N
_{0}e^{-λt} - t
_{av}= (λ/N_{0}) ∫tN_{0}e^{-λt}dt = (λ/N_{0})∫t e^{-λt}dt - t
_{av}= λ_{0}^{∞}∫te^{-λt}dt - =λ[t∫e
^{-λt}dt -∫(∫e^{-λt}.1)dt - =λ[(te
^{-λt})/(-λ)]_{0}^{∞}-λ [(e^{-λt})/(-λ) dt]_{0}^{∞} - =λ[0] –λ[(e
^{-λt})/λ^{2}]_{0}^{∞} **t**_{av}= (1/ λ)

- Let number of radioactive sample at t=0 =N

- Using t
_{(1/2)}= (0.693/λ) and t_{av}= (1/ λ) **t**_{av}= (t_{(1/2)}/0.693)

__Problem:-__

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value?

__Answer:-__

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N_{0}

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of N_{0} remains after decay. Hence, we can write:

(N/N_{0})= 3.125 %=( 3.125)/ (100) = (1/32)

But (N/N_{0}) =e^{-}^{ λt}

Where, λ = Decay, constant t = Time

Therefore e^{- (}^{λt)} = (1/32)

- (λt) = ln (1) – ln (32)

- (λt) = 0 -3.4657

t= (3.4657)/ (λ)

Since λ = (0.693)/ (T)

Therefore t= (3.466)/ ((0.693)/ (T)) ≈ 5T years.

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of N_{0} remains after decay. Hence, we can write:

(N/N_{0})=1 %=( 1/100)

But (N/N_{0}) =e^{-}^{ λt}

Therefore e^{- (}^{λt)} = (1/100)

- (λt) = ln (1) – ln (100)

- (λt) = 0 - 4.6052

t= (4.6052)/ (λ)

Since λ = (0.693)/ (T)

Therefore t= (4.6052)/ ((0.693)/ (T)) =6.645T years.

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

** Problem:- **Obtain the amount (

** Answer:- **The strength of the radioactive source is given as:

(dN/dt) =8.0 mCi

=8x10^{-3}x 3.7x10^{10}

=29.6x10^{7} decay/s.

Where,

N = Required number of atoms

Half-life of (^{60}_{27}Co), T_{1/2} = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10^{8} s

For decay constant λ, we have the rate of decay as:

(dN/dt) = (λN)

Where λ = (0.693/T_{1/2}) = (0.693)/ (1.67x10^{8}) s^{-1}

Therefore N = (1/λ) (dN/dt)

= (29.6x10^{7})/ (0.693)/ (1.67x10^{8}) =7.133 x10^{16} atoms

For (^{60}_{27}Co):

Mass of 6.023 × 10^{23} (Avogadro’s number) atoms = 60 g

Therefore, Mass of (7.133 x 10^{16}) atoms= (60x7.1333x10^{16})/ (6.023x10^{23})

=7.106x10^{-6}g

Hence, the amount of (^{60}_{27}Co) necessary for the purpose is 7.106 × 10^{−6} g.

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