I E Irodov Solutions for Chapter: ATOMIC AND NUCLEAR PHYSICS, Exercise 5: RADIOACTIVITY

Find the amount of heat generated by $1.00 \mathrm{mg}$ of ${\mathrm{Po}}^{210}$ preparation during the mean lifetime period of these nuclei, if the emitted alpha-particles are known to possess the kinetic energy $5.3 \mathrm{MeV}$and practically all daughter nuclei are formed directly in the ground state.

Avogadro constant $N_A$$=6.023\times {10}^{23} {\mathrm{mol}}^{-1}$.

The alpha-decay of ${\mathrm{Po}}^{210}$ nuclei (in the ground state) is accompanied by the emission of two groups of alpha-particles with kinetic energies $5.30$ and $4.50 \mathrm{MeV}.$ Following the emission of these particles, the daughter nuclei are found in the ground and excited states. Find the energy of gamma-quanta emitted by the excited nuclei.

Find the energy $Q$ liberated in ${\beta }^{-}$ and ${\beta }^{+}$ decays and in $K$ capture, if the masses of the parent atom $M_p,$ the daughter atom $M_d$ and an electron $m$ are known.( $c$ is speed of light)

Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta particles emitted by ${\mathrm{Be}}^{10}$ nuclei and the corresponding kinetic energy of the recoiling daughter nuclei formed directly in the ground state.

${\mathrm{Be}}^{10}=10.016711 \mathrm{amu}$
${\mathrm{B}}^{10}=10.016114 \mathrm{amu}$

Evaluate the amount of heat produced during a day by ${\beta }^{-}$ active ${\mathrm{Na}}^{24}$ preparation of mass $m=1.0 \mathrm{mg}$. The beta-particles are assumed to possess an average kinetic energy equal to $1/3$ of the highest possible energy of the given decay. The half-life of ${\mathrm{Na}}^{24}$ is $T=15$ hours.

Mass of ${\mathrm{Na}}^{24}$ is $24.00903 \mathrm{amu}$ and of ${\mathrm{Mg}}^{24}$ is $24.01496 \mathrm{amu}$ , energy equivalent of neutrinos or electron is $0.511 \mathrm{MeV}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy, speed of light $3\times {10}^8 {\mathrm{ms}}^{-1}$.

Taking the values of the atomic masses from the tables, calculate the kinetic energies of a positron and a neutrino emitted by ${\mathrm{C}}^{11}$ nucleus for the case, when the daughter nucleus does not recoil.

Mass defect due to positron is $0.00213 \mathrm{amu}$, energy equivalent of neutrinos or electron is $0.511 \mathrm{MeV}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931.5 \mathrm{MeV}$ of energy, speed of light $3\times {10}^8 {\mathrm{ms}}^{-1}$.

Find the kinetic energy of the recoil nucleus in the positron decay of ${\mathrm{N}}^{13}$ nucleus for the case when the energy of the positron is maximum.

energy equivalent of neutrinos or electron is $0.511 \mathrm{MeV}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931 \mathrm{MeV}$ of energy, speed of light $3\times {10}^8 {\mathrm{ms}}^{-1}$.

From the tables of atomic masses, determine the velocity of a nucleus appearing as a result of $K$ capture in a ${\mathrm{Be}}^7$ atom, provided the daughter nucleus turns out to be in the ground state.

energy equivalent of neutrinos or electron is $0.511 \mathrm{MeV}$

$1 \mathrm{amu}=1.67\times {10}^{-24} \mathrm{g}$ and $1 \mathrm{amu}$ is equivalent to $931 \mathrm{MeV}$ of energy, speed of light $3\times {10}^8 {\mathrm{ms}}^{-1}$.