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प्रश्न
Derive the mathematical expression for law of radioactive decay for a sample of a radioactive nucleus
Deduce the expression, N = N0 e−λt, for the law of radioactive decay.
Derive the expression N = Noe-λt where symbols have their usual meanings
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उत्तर
In any radioactive sample which undergoes α, β or γ-decay, it is found that the number of nuclei undergoing decay per unit time is proportional to the total number of nuclei in the sample.
If N is the number of nuclei in the sample and ΔN undergoes decay in time Δt, then we have
`(DeltaN)/(Deltat) propN`
`:.(DeltaN)/(Deltat)=lambdaN`
where λ is called the radioactive decay constant or disintegration constant.
The change in the number of nuclei in the sample is dN = −ΔN in time Δt, i.e. in the limit dt → 0. Thus, the rate of change of N is
`(dN)/dt= -lambdaN`
`:.(dN)/N = -lambdadt`
Now, integrating both the sides, we get
`int_(N_0)^N (dN)/N=-lambdaint_(t_0)^tdt`
∴ lnN - lnN0 = -λ(t-t0)
Here, N0 is the number of radioactive nuclei in the sample at some arbitrary time t0 and N is the number of radioactive nuclei at any subsequent time t. Setting t0 = 0 and rearranging, we get
`ln`
∴ N(t) = N0e-λt
This is the law of radioactive decay.
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संबंधित प्रश्न
(a) Write the basic nuclear process involved in the emission of β+ in a symbolic form, by a radioactive nucleus.
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(i)`""_16^11C->_y^zB+x+v`
(ii)`""_6^12C+_6^12C->_a^20 Ne + _b^c He`
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\[\ce{^11_6C -> ^11_5B + e+ + \text{v}}\] : T1/2 = 20.3 min
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values: `"m"(""_6^11"C") = 11.011434 u and "m"(""_6^11"B") = 11.009305 "u"`
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\[\ce{^1_1H + ^3_1H -> ^2_1H + ^2_1H}\]
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`"m"(""_1^3"H")` = 3.016049 u
`"m"(""_6^12"C")` = 12.000000 u
`"m"(""_10^20"Ne")` = 19.992439 u
The Q value of a nuclear reaction A + b → C + d is defined by
Q = [mA+ mb − mC − md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
\[\ce{^12_6C + ^12_6C ->^20_10Ne + ^4_2He}\]
Atomic masses are given to be
`"m"(""_1^2"H")` = 2.014102 u
`"m"(""_1^3"H")` = 3.016049 u
`"m"(""_6^12C)` = 12.000000 u
`"m"(""_10^20"Ne")` = 19.992439 u
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