JEE Physics Nuclei Notes Set 04

Physics Concept Notes for JEE: Nuclei

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Nuclear Size, Mass and Binding Energy

1. Radius \( R=R_0A^{1/3} \) with \( R_0\approx1.2 \) fm. Volume is proportional to A, so the density is nearly the same for all nuclei.
2. Mass defect \( \Delta m=Zm_p+(A-Z)m_n-M \) is positive for every stable nucleus. Binding energy is \( \Delta m\,c^2 \), and 1 u corresponds to 931 MeV.
3. Binding energy per nucleon rises with A to a maximum near iron (about 8.7 MeV at A = 56) and then falls slowly. This is why fusion of light nuclei and fission of heavy nuclei both release energy.
4. Energy released in a nuclear reaction is the difference of the total binding energies: for two deuterons forming He-4, \( 28-2\times2.2=23.6 \) MeV.

Radioactive Decay

QuantityFormula
Decay law\( N=N_0e^{-\lambda t}=N_0\left(\tfrac12\right)^{t/T_{1/2}} \)
Half-life and mean life\( T_{1/2}=\dfrac{0.693}{\lambda},\ \tau=\dfrac{1}{\lambda} \)
Activity\( A=\lambda N=A_0e^{-\lambda t} \)
Two decay modes\( \lambda=\lambda_1+\lambda_2 \), so \( \dfrac1\tau=\dfrac1{\tau_1}+\dfrac1{\tau_2} \)

1. After n half-lives the fraction left is \( \tfrac1{2^n} \). To reduce to 1 percent takes \( \log_2 100 \approx 6.64 \) half-lives.
2. Alpha decay lowers A by 4 and Z by 2. Beta-minus decay raises Z by 1 with A unchanged. Gamma emission changes neither A nor Z.
3. For a chain from mass number \( A_1 \) to \( A_2 \): number of alpha particles \( =\dfrac{A_1-A_2}{4} \), then beta particles are found from the change in Z.
4. In alpha decay of a nucleus at rest, the momenta are equal and opposite, so the kinetic energies are in the ratio of the inverse masses: the alpha particle gets \( \dfrac{A-4}{A} \) of the energy released.
5. The continuous energy spectrum of beta particles (zero up to a maximum \( E_0 \)) led to the neutrino hypothesis.

Fission, Fusion and Reactors

1. Fission fragments are neutron rich and undergo beta-minus decay. Neutrons released can start a chain reaction.
2. In a reactor, a moderator (heavy water, graphite) slows neutrons down, control rods (cadmium, boron) absorb neutrons to keep the chain reaction steady, and a coolant removes the heat.
3. Fusion needs extremely high temperatures so that nuclei overcome their electrical repulsion, as in the Sun.

Nuclei Concepts and Summary for JEE Physics

Key Concepts and Summary for JEE Physics Nuclei

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