Embibe Experts Solutions for Chapter: Nuclear Physics, Exercise 4: Exercise - 4

(a) Find the energy needed to remove a neutron from the nucleus of the calcium isotope ${}_{20}{}^{42}\mathrm{Ca}$
(b) Find the energy needed to remove a proton from this nucleus
(c) Why are these energies different?

Atomic masses of ${}_{20}{}^{41}\mathrm{Ca}$ and ${}_{20}{}^{42}\mathrm{Ca}$ are $40.962278\mathrm{u}$ and $41.958622\mathrm{u}$ respectively.
Atomic mass of ${}_{19}{}^{41}\mathrm{K}$ is $40.961825\mathrm{u}$, Mass of Proton is $1.007826\mathrm{u}$

A nucleus $\mathrm{X}$, initially at rest, undergoes alpha-decay according to the equation.

${}_{92}{}^{\mathrm{A}}\mathrm{X} \to + {}_{\mathrm{Z}}{}^{228}\mathrm{Y} + \mathrm{\alpha }$

(a) Find the values of $A$ and $Z$ in the above process.
(b) The alpha particle produced in the above process is found to move in a circular track of radius $0.11\mathrm{m}$ in a uniform magnetic field of $3\mathrm{tesla}$. Find the energy (in $\mathrm{MeV}$) released during the process and the binding energy of the parent nucleus $\mathrm{X}$.

Given that $m\left(\mathrm{Y}\right)=228.03\mathrm{u}; m( {}_0{}^1\mathrm{n} )=1.009\mathrm{u}.$
$m\left({}_2{}^4\mathrm{He}\right)=4.003\mathrm{u} ; m( {}_1{}^1\mathrm{H} )=1.008\mathrm{u}.$

$100\mathrm{millicuries}$ of radon which emits $5.5\mathrm{MeV} \alpha$-particles are contained in a glass capillary tube $5\mathrm{cm}$ long with internal and external diameters $2$ and $6\mathrm{mm}$ respectively Neglecting and effects and assuming that the inside of the tube is uniformly irradiated by the particles which are stopped at the surface calculate the temperature difference between the walls of a tube when steady thermal conditions have been reached.
Thermal conductivity of glass $= 0.025\mathrm{Cal} {\mathrm{cm}}^{-2} {\mathrm{s}}^{-1} {\mathrm{C}}^{-1}$
$\mathrm{Curie}=3.7 \times {10}^{10}\mathrm{disintergration} \mathrm{per} \mathrm{second}$
$\mathrm{J} = 4.18 \mathrm{Joule} {\mathrm{Cal}}^{-1}$

Radium being a member of the uranium series occurs in uranium ores. If the half lives of uranium and radium are respectively $4.5 \times {10}^9$ and $1620\mathrm{years}$ calculate the $\frac{{\mathrm{N}}_{\mathrm{radiun}}}{{\mathrm{N}}_{\mathrm{Uranium}}}$ in Uranium ore at equilibrium.

${}^{90}\mathrm{Sr}$ decays to ${}^{90}\mathrm{Y}$ by $\mathrm{\beta }$ decay with a half-life of $28 \mathrm{years}$. ${}^{90}\mathrm{Y}$ decays by $\mathrm{\beta }$ decay to ${}^{90}\mathrm{Zr}$ with a half-life of $64 \mathrm{h}$. A pure sample of ${}^{90}\mathrm{Sr}$ is allowed to decay. What is the value of $\frac{{\mathrm{N}}_{\mathrm{Sr}}}{{\mathrm{N}}_{\mathrm{Y}}}$ after $\left(a\right)$ $1 \mathrm{h}$ $\left(b\right)$ $10 \mathrm{years}$?

The element Curium ${}_{96}{}^{248}\mathrm{Cm}$ has a mean life of ${10}^{13}\mathrm{seconds}$. Its primary decay modes are spontaneous fission and $\mathrm{\alpha }$-decay, the former with a probability of$8\%$ and the latter with a probability of $92\%$. Each fission releases $200\mathrm{MeV}$ of energy. The masses involved in $\mathrm{\alpha }$-decay are as follows :
atomic masses of atoms are ${}_{96}{}^{248}\mathrm{Cm}=248.072220\mathrm{u}, {}_2{}^4\mathrm{He}=4.002603\mathrm{u} \& {}_{94}{}^{244}\mathrm{Pu}=244.064100\mathrm{u} (1\mathrm{u}=931\mathrm{MeV}/{\mathrm{c}}^2)$. Calculate the power output from a sample of ${10}^{20} \mathrm{Cm}$ atoms.

Nucleus ${}_3\mathrm{A}^7$ has binding energy per nucleon of $10 \mathrm{MeV}$. It absorbs a proton and its mass increases by $\frac{99}{100}$ times the mass of proton. Find the new binding energy of the nucleus so formed [Take energy equivalent of proton $= 930 \mathrm{MeV}$]

The nucleus of ${}_{90}{}^{230}\mathrm{Th}$ is unstable against $\mathrm{\alpha }$-decay with a half-life of $7.6 \times {10}^3 \mathrm{years}$. Write down the equation of the decay and estimate the kinetic energy of the emitted $\mathrm{\alpha }$-particle from the following data.
$\mathrm{m}\left({}_{90}{}^{230}\mathrm{Th}\right) =230.0381 \mathrm{amu}$, $\mathrm{m}\left({}_{88}{}^{226}\mathrm{Ra}\right) = 226.0254 \mathrm{amu}$ and $\mathrm{m}\left({}_2{}^4\mathrm{He}\right) = 4.0026 \mathrm{amu}$.