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Q.Derive an expression for average life of a radioactive sample. OR Distinguish between Nuclear fission and Nuclear fusion.

Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2021Subjective· 3mImportance★★★★★
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Average life is the reciprocal of the decay constant, obtained by averaging the lifetimes of all nuclei in a sample using the exponential decay law.

Average life derivation:

For a radioactive sample obeying N=N0e−λtN=N_0e^{-\lambda t}, the number of nuclei that decay in the interval tt to t+dtt+dt is

dN=λN0e−λt dtdN = \lambda N_0 e^{-\lambda t}\,dt

Each of these nuclei has "lived" for a time tt. So the average (mean) life is the total lifetime of all nuclei divided by the total number of nuclei N0N_0:

τ=∫0∞t λN0e−λt dtN0=λ∫0∞t e−λt dt\tau = \frac{\displaystyle\int_0^\infty t\,\lambda N_0 e^{-\lambda t}\,dt}{N_0} = \lambda\int_0^\infty t\,e^{-\lambda t}\,dt

Using the standard integral ∫0∞t e−λtdt=1/λ2\int_0^\infty t\,e^{-\lambda t}dt = 1/\lambda^2,

τ=λ⋅1λ2=1λ\tau = \lambda\cdot\frac{1}{\lambda^2} = \frac{1}{\lambda}

Since half-life T1/2=ln⁡2λT_{1/2} = \dfrac{\ln 2}{\lambda}, we also get τ=T1/2ln⁡2≈1.44 T1/2\tau = \dfrac{T_{1/2}}{\ln 2} \approx 1.44\,T_{1/2}.

OR — Nuclear fission vs nuclear fusion:

Nuclear FissionNuclear Fusion
A heavy nucleus (e.g. 92235U^{235}_{92}\text{U}) splits into two lighter nuclei, usually on absorbing a neutronTwo light nuclei (e.g. isotopes of hydrogen) combine to form a heavier nucleus
Can be initiated at ordinary temperature (by a slow neutron)Requires extremely high temperature (~10710^7 K) and pressure to overcome Coulomb repulsion

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