Class 12 Physics Nuclei Terminologies related to Radioactive decay

1. Decay Rate: -
1. Decay rate is defined as the number of nuclei decaying per unit time.
2. Denoted by ‘R’.
3. 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 = N0 e-λt, therefore
• R=(d/dt)( N0 e-λt)
• =- N0 [e-λt(-λ)]
• R= N0λ e-λt ;
• At t=0 , R0= N0λ (equation (1))
• Therefore R=R0 e-λt=>decay constant also changes exponentially.
1. Activity of radioactive sample is defined as the total decay rate R of a sample of one or more radionuclides.
2. Decay rate of whole sample is considered.
3. I. unit: - Becquerel(Bq); 1Bq = (1 decay/sec).
4. Other unit: - Curie(Ci); 1C i=3.7 x1010
1. Half – life of radioactive sample is defined as the time at which number of nuclei reduces to one-half of their initial values.
2. It is denoted by t (1/2).
3. Half-life tells how long radioactive nuclei can last.
4. Mathematical expression:-
• From radioactive law , N=N0e-λt (equation(1))
• Initially number of sample = N0.
• Later number of samples will become=(N0/2).
• Therefore from equation(1)- (N0/2) =N0e-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)
1. Mean life is defined as the average life of a nuclei in the radioactive sample.
2. Any nuclei can decay at any interval of time.
3. It is denoted by tav.
4. Mathematically:-
• Let number of radioactive sample at t=0 =N0.
• 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 =tav= (0∫(tλN)dt/N0)
• = (λ/N0) 0∫ (tN) dt.
• From exponential law of decay :- N =N0 e-λt
• tav= (λ/N0) ∫tN0 e-λt dt = (λ/N0)∫t e-λt dt
• tav = λ0∫te-λtdt
• =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)/λ2]0
• tav = (1/ λ)
• Using t(1/2) = (0.693/λ)  and tav = (1/ λ)
• tav = (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?

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N0

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

It is given that only 3.125% of N0 remains after decay. Hence, we can write:

(N/N0)= 3.125 %=( 3.125)/ (100) = (1/32)

But (N/N0) =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 N0 remains after decay. Hence, we can write:

(N/N0)=1 %=( 1/100)

But (N/N0) =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 (6027Co) of necessary to provide a radioactive source of 8.0 mCistrength. The half-life of (6027Co) is 5.3 years.

(dN/dt) =8.0 mCi

=8x10-3x 3.7x1010

=29.6x107 decay/s.

Where,

N = Required number of atoms

Half-life of (6027Co), T1/2 = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 108 s

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

(dN/dt) = (λN)

Where λ = (0.693/T1/2) = (0.693)/ (1.67x108) s-1

Therefore N = (1/λ) (dN/dt)

= (29.6x107)/ (0.693)/ (1.67x108) =7.133 x1016 atoms

For (6027Co):

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

Therefore, Mass of (7.133 x 1016) atoms= (60x7.1333x1016)/ (6.023x1023)

=7.106x10-6g

Hence, the amount of (6027Co) necessary for the purpose is 7.106 × 10−6 g.

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