Atomic Nucleus
Hydrogen Atom
Positive charge proton
$m_p \approx 2000 m_e$
$|e_p| = |q_e|$
Atom with Z Electrons
There are Z number of protons in its nucleus.
$4{He}_2=Z$
$Z=2$
Protons
$m_{H e} ≄ 2 m_H$
$\simeq 4 m_H$
Experiments of Chadwick
Bombard light elements such as Boron and Beryllium with $\alpha$ particles
$\alpha$ particles (were obtained from the radioactive decay of Polonium. Energies: $\sim 5 {MeV}$
Experiments Results
Protons and neutral radiation
Neutral particles had great penetrating power (greater than X-Rays)
The neutral radiation could further eject electrons from a number of protons.
Neutron
Mary and Pierre Curie: Photons (Compton scattering analogy)
Chadwick: They were new kind of particles with mass almost same as of protons (Energy Momentum conservation)
$M_n \approx 1.15 M_p$ (Chadwick’s estimate)
$M_n=1.001 M_p$ (Current Value)
An atomic Nucleus consists of
Protons (Positive Charge)
Neutrons (Neutral)
$M_p \approx M_n$
There is no evidence for existence of electrons in the nucleus
There are always as many protons as there are electrons orbiting the nucleus
$\left| \frac{M_n-M_p}{M_p}\right| \simeq 10^{-3}$
Periodic Properties
Big question:- What about the number of neutrons?
What are isotopes, isobars and isotones?
Matters of jargon
Periodic Table of the Elements
Notation
${ }_Z^A X$
X: Nucleus (Hydrogen, Carbon, Phosphorus, …
A: Total Number of protons + neutrons (Atomic Weight) (Total number of nucleons)
Z: Total number of protons (Atomic Number)
A-Z: Total number of neutrons
Nuclide: A nucleus with both $A$ and $Z$ fixed
Definition
Two nuclides with the same Z are called isotopes
Two nuclides with the same A are called isobars
Two nuclides with the same A-Z are called isotones
Isotopes
${ }_1^1 {H},{ }_1^2 {H},{ }_1^3 {H}$
Hydrogen, Deuterium and Tritium
Similarly Helium and Tritium for Isobars
Isobar
$^3 _2He$
A = 3
Z = 2
$^3 _1H$
Z = 1
Size of the Nucleus of an Atom
Rough scale :-
$r_{{nucleus}} \sim 10^{-14}-10^{-15} m $
Charge Distribution
$ \text{Charge density} :1 fm = 10^{-15} m $
Sizes of Nuclei 1
Scattering
Nuclear Sizes
Two parameters: A and Z
Depends only on A:
$R = R_0 A^{1/3}$
$M = M_0 A$
$V = V_0 A$
$R_0=1.25 \times 10^{-15} \mathrm{m} ; \mathrm{R}_{{p}} \approx 0.85 \times 10^{-15} {m}$
$R(A)=R_0 A^{1/3}$
$M = M_0 V$
$M = {\rho_x} V=\rho_{{const.}} \times R^3(A)$
$\simeq(P_c R_0^3) A$
$M_0$
Given Radius of $^{16}O$ We could be able to get $R(^{12}C)$
$R(^{16}O) = R_0(16)^{1/3}$
$R(^{12}C) = R_0(12)^{1/3}$
$\frac{R(^{12}C)}{R(^{16}O)} = (\frac{12}{16})^{1/3}$
$R(^{12}C) = (\frac{3}{4})^{1/3} R(^{16}O)$
$\simeq 1$
Nucleus
$\simeq 10^{-15}$ m
$R \sim R_0 A^{1/3}$
Sensitive to changes in the scale of f m
Masses of Nuclei Relativity
Newtonian case
P = mv ; $E = \frac{1}{2} m v^2$
v = at
v > c
More on Nuclear Masses
Mass energy relation
No material particle can move with a speed $v \geq c$
There is no restriction on the energy a body can possess
Therefore, $p \neq m v, E \neq \frac{1}{2} m v^2$
Inertia depends on the speed!
Force
$\frac{d p}{d t}=m \frac{d v}{d t}=F$
$a=\frac{F}{m}$
independent of speed
integration
$v(t)=(\frac{F}{m}) {t}+v_0>C$
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Mass Energy Equivalence
Einstein relation
$M(v) = M_0 \frac{1}{\sqrt{(1-v^2 / c^2)}}$
$E =M(v) c^2$
p(v) = M(v)v
$E^2 =p^2 c^2+M_0^2 c^4$
$M(v)=M_0 \frac{1}{\sqrt{1-v^2 / c^2}}$
$\frac{v}{c} \ll 1$
$M(v) \simeq M_0[1+\frac{1}{2}(\frac{v^2}{c^2})]$
$\simeq M_0 ~ small ~\frac{v}{c}$
$\vec{p}(v)={m(v)} \vec{v}$
$\frac{d \vec{p}}{d t}=\vec{F}=\text { const. }$
$\vec{p}(t)=\vec{F} \times t$;
$=m(v(t)) \vec{v}(t)$
F(t) = constant
$\vec{p}(t)=\vec{F} \times t$
$\vec{E}(v) =\frac{m_0 c^2}{\sqrt{1-v^2 / c^2}}$
$=m(v) c^2$
$\simeq m_0 c^2[1+\frac{1}{2} \frac{v^2}{c^2}+\cdots]$
$=m_0 c^2+\frac{1}{2} m_0 v^2$
$E^2=p^2 c^2+M_0^2 c^4$
has non-trivial solutions even when $M_v=0$
E = p c ;
$v=\frac{d \varepsilon}{d p}=C$
Mass Speed Relation
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