Atomic Structure [Normal Revision ]

 

Atomic Structure (Physics)

Thomson's Model

• Thomson's tried to explain about the arrangement of +ve & -ve charges in an atom.
  
  
• An atom is a +vely charged sphere of diameter about 10−10${}^{-10}$m is which +ve charge are uniformly distributed where as -ve charges are slightly embedded like plum in a pudding.
  
  
• This model is also called plum-pudding model.
  
  
• He explained unequal concentration of electric change.
  
  
• Atom is electrically neutral, the concentration of electric charge must be equal.
  
  
• This model could not explain about the atomic stability.
  
  
• This model could not explain about 
  $\alpha$ -particle scattering experiment.
  Hence, this model is fails to exist

Rutherford's Model

Rutherford's carried out of α$\alpha$ particle scattering experiment and achieved the following observation.

Observation

1. Most of α$\alpha$ -particles cross thin gold foil without any deflection.
2. Some of α$\alpha$ -particles are deflected with particular angle i.e. less than 900${}^0$i.e. acute angle.
3. Few α$\alpha$ -particles are totally back at an angle 1800${}^0$.
   
   

Postulates

1. The +vely charged and massive particles are concentrated to the inner core of an atom that inner core is called nucleus having diameter about 10−14${}^{-14}$m to 10−15${}^{-15}$m.
2. The space around the nucleus is totally empty.
3. The -vely charged particles are revolving round the circular path, known as electrons.
   
   
   

Failures

1. This model could not explain atomic stability.
2. This model could not explain origin of line spectra.
   
   Therefore, Rutherford's model fails to exist

Application of Rutherford model is to determine the closed distance of approach

When +vely charged particle approaches towards a stationary nucleus then due to repulsion between them the kinetic energy of +vely charged particle gradually decreases & a stage come at which potential energy gain by charged particle due to stationary nucleus balanced its K.E.

We have,

P.E=K.E$\mathrm{P}.\mathrm{E}=\mathrm{K}.\mathrm{E}$

q1q24πϵoro=K.E$\frac{q_1{q}_2}{4\pi {ϵ}_o{r}_{\mathrm{o }}}= \mathrm{K}.\mathrm{E}$

KZ1eZ2ero=K.E$\mathrm{K}\frac{{\mathrm{Z}}_1\mathrm{e}{\mathrm{Z}}_2\mathrm{e}}{{\mathrm{r}}_o}=\mathrm{K}.\mathrm{E}$

Closest distance (ro)=kZleZ2e2K.E$\left({\mathrm{r}}_{\mathrm{o}}\right)=\frac{k{Z}_{\mathrm{l}}e{Z}_2{e}^2}{K.E}$

∴$\therefore$ The closest distance of approach of any +vely charged particle to the stationary nucleus.

Bohr's Model

• Electron revolves round the stationary orbit. The necessary centripetal force provided by the electrostatic force of attraction between electron & nucleus.
  
  i.e mv2r=kze2r2⋯⋯(i) $\frac{m{v}^2}{r}=\mathrm{k}\frac{{\mathrm{z}\mathrm{e}}^2}{r^2}\cdots \cdots (\text{i) }$
  
  
• Those orbits are called stationary orbits in which an angular momentum of an e−${}^{-}$ is integral multiple of h/2π$\pi$.
  
  i.e. mvr=nh2π⋯⋯(ii) $\mathrm{mvr}=\frac{nh}{2\pi }\cdots \cdots (\text{ii) }$
  
  
• Electrons absorbs energy to jump from lower orbit to higher orbit and when an electron jump from higher energy state to lower energy state, it will radiate energy.
  
  i.e ΔE=En2−En1$\mathrm{\Delta }E=E_{n_2}-E_{n_1}$
  
  hc/λ=En2−En1$hc/\lambda =E_{n_2}-E_{n_1}$
  
  Radiated wavelength is:
  
  (λ)=hcEn2−En1⋯⋯(iii) $(\lambda )=\frac{hc}{E_{n_2}-E_{n_1}}\cdots \cdots (\text{iii) }$
  
  
• The no of possible photons due to transition of electron from energy level are:
  
  N=n(n−1)2⋯⋯(iv) $N=\frac{n(n-1)}{2}\cdots \cdots (\text{iv) }$
  
  where, n is the no. of orbit.

• Radius of nth$n^{th}$ orbit

  The radius of nt orbit of hydrogen like atom is:rn=n2h24π2kzme2×n2z=0.53×10−10n2z$\begin{array}{rl}{r}_n& =\frac{n^2{h}^2}{4{\pi }^2\mathrm{kzm}{e}^2}\times \frac{n^2}{z}\\ & =0.53\times {10}^{-10}\frac{n^2}{z}\end{array}$
  
  ∴rn∝n2z$\therefore r_n\propto \frac{n^2}{z}$
  
  for H−atom,z=1$\mathrm{H}-\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m},\mathrm{z}=1$ so, r=ε0h2πme2n2$\mathrm{r}=\frac{{\epsilon }_0{\mathrm{h}}^2}{\pi {\mathrm{m}\mathrm{e}}^2}{\mathrm{n}}^2$
  
  If n=1,r1=0.53×10−10m$\mathrm{n}=1,{\mathrm{r}}_1=0.53\times {10}^{-10}\mathrm{m}$
  =0.53A˚$=0.53\mathring{A}$
  
  if n=2,r2=2.12A˚$\mathrm{n}=2,{\mathrm{r}}_2=2.12\mathring{A}$ and so on.
  
  
  
  

• Velocity of e−$e^{-}$ in nth$n^{th}$ orbit

  Velocity of e−$e^{-}$ in nth$n^{th}$ orbit of atom is:
  
  We have,
  
  mvnrn=nh2π$m{v}_n{r}_n=\frac{nh}{2\pi }$
  
  Vn==Vn=Vn=∴Vn∝znnh2πmrn=e22ε0hznn×6.64×10−342×3.14×9.1×10−31×0.53×10−10n/zC137zn2.18×106zn$\begin{array}{rl}{\mathrm{V}}_{\mathrm{n}}=& \frac{nh}{2\pi m{r}_n}=\frac{{\mathrm{e}}^2}{2{\epsilon }_0\mathrm{h}}\frac{\mathrm{z}}{\mathrm{n}}\\ =& \frac{n\times 6.64\times {10}^{-34}}{2\times 3.14\times 9.1\times {10}^{-31}\times 0.53\times {10}^{-10}\mathrm{n}/\mathrm{z}}\\ {\mathrm{V}}_{\mathrm{n}}=& \frac{\mathrm{C}}{137}\frac{\mathrm{z}}{\mathrm{n}}\\ {\mathrm{V}}_{\mathrm{n}}=& 2.18\times {10}^6\frac{\mathrm{z}}{\mathrm{n}}\\ \therefore {\mathrm{V}}_{\mathrm{n}}\propto \frac{\mathrm{z}}{n}\end{array}$
  
  For hydrogen atom Z=1$Z=1$. So, v=C1371n$v=\frac{C}{137}\frac{1}{n}$
  
  If n=1,v1=C137$n=1,v_1=\frac{C}{137}$
  
  If n=2,v2=C13712$n=2,v_2=\frac{C}{137}\frac{1}{2}$
  
  & so on.
  
  
  
  

• Energy in nth$n^{th}$ orbit

  Total energy in nth$n^{th}$ orbit is:
  
  En=K.E.+P.E.$E_n=K.E.+P.E.$
  
  Again,
  
  P.E.=−Ze24πε0r$\mathrm{P}.\mathrm{E}.=\frac{-{\mathrm{Z}\mathrm{e}}^2}{4\pi {\epsilon }_0\mathrm{r}}$
  
  K.E.=Ze28πε0r$\mathrm{K}.\mathrm{E}.=\frac{{\mathrm{Z}\mathrm{e}}^2}{8\pi {\epsilon }_0\mathrm{r}}$
  
  ∴En=−Ze28πε0r=−Z2me48ε20n2h2$\therefore {\mathrm{E}}_{\mathrm{n}}=-\frac{{\mathrm{Z}\mathrm{e}}^2}{8\pi {\epsilon }_0\mathrm{r}}=-\frac{{\mathrm{Z}}^2{\mathrm{m}\mathrm{e}}^4}{8{\epsilon }_0^2{\mathrm{n}}^2{\mathrm{h}}^2}$
  
  =−13.6z2n2ev$=-13.6\frac{{\mathrm{z}}^2}{n^2}\mathrm{e}\mathrm{v}$
  
  ∴En∝z2n2$\therefore {\mathrm{E}}_{\mathrm{n}}\propto \frac{{\mathrm{z}}^2}{n^2}$
  
  For hydrogen Z=1$Z=1$, So, En=−13.61n2ev$E_n=-13.6\frac{1}{n^2}ev$
  
  
  

• Time period for an e−$e^{-}$ in the nth$n^{th}$ orbit

  Tn=2πrnVn=2π×0.53×10−10n2/z2.18×106z/n${\mathrm{T}}_{\mathrm{n}}=\frac{2\pi r_n}{V_n}=\frac{2\pi \times 0.53\times {10}^{-10}{n}^2/\mathrm{z}}{2.18\times {10}^6\mathrm{z}/\mathrm{n}}$
  
  Tn∝n3z2${\mathrm{T}}_{\mathrm{n}}\propto \frac{n^3}{{\mathrm{z}}^2}$
  
  For hydrogen z=1so,Tn∝n2$\mathrm{z}=1\mathrm{s}\mathrm{o},{\mathrm{T}}_{\mathrm{n}}\propto {\mathrm{n}}^2$
  
  
  
  

• Frequency of an e−$e^{-}$ in the nth$n^{th}$ orbit

  fn=Vn2πrn${\mathrm{f}}_{\mathrm{n}}=\frac{V_n}{2\pi r_n}$
  
  fn∝z2n3${\mathrm{f}}_{\mathrm{n}}\propto \frac{z^2}{n^3}$
  
  For hydrogen z=1$\mathrm{z}=1$ So, fn∝1n3${\mathrm{f}}_{\mathrm{n}}\propto \frac{1}{n^3}$
  
  Eqn for wavelength is, 1λ=R(1n21−1n22)$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
  
  Where,
  
  R=me48ε20ch3=Rydberg's constant=1.097×107m−1$\begin{array}{rl}R=\frac{m{e}^4}{8{\epsilon }_0^{2}c{h}^3}& =\text{Rydberg's constant}\\ & =1.097\times {10}^7{m}^{-1}\end{array}$
  
  n1=$n_1=$ lower energy state
  n2=$n_2=$ higher energy state
  
  

  • Current :
    

    In=qt=qf${\mathrm{I}}_{\mathrm{n}}=\frac{q}{t}=\mathrm{q}\mathrm{f}$
    
    ∴In∝z2n3$\therefore {\mathrm{I}}_{\mathrm{n}}\propto \frac{{\mathrm{z}}^2}{n^3}$
    
    
    

  • Magnetic field at the center is :
    

    Bn=μoI2r${\mathrm{B}}_{\mathrm{n}}=\frac{{\mu }_{o}I}{2r}$
    
    ∴Bn∝z2/n3n2/z∝z3n5$\therefore {\mathrm{B}}_{\mathrm{n}}\propto \frac{{\mathrm{z}}^2/{\mathrm{n}}^3}{{\mathrm{n}}^2/\mathrm{z}}\propto \frac{{\mathrm{z}}^3}{n^5}$

Emission Spectra

• Lyman Series:

  If n1=1&n2=2,3,4,…$n_1=1\mathrm{\&}{n}_2=2,3,4,\dots$
  
  Then 1/λ=R[1n21−1n22]$1/\lambda =\mathrm{R}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
  
  For 1st $1^{\text{st }}$ line n1=1&n2=2${\mathrm{n}}_1=1\mathrm{\&}{\mathrm{n}}_2=2$
  
  so, 1λ=R[112−122]$\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$
  
  for series limit n1=1,n2=∞${\mathrm{n}}_1=1,{\mathrm{n}}_2=\mathrm{\infty }$
  
  so, 1λ=R$\frac{1}{\lambda }=\mathrm{R}$
  
  →$\to$ It lies in u-v region
  
  
  

• Balmer series:

  If n1=2&n2=3,4,5,…$n_1=2\mathrm{\&}{n}_2=3,4,5,\dots$
  
  So, 1/λ=R[122−1n22]$1/\lambda =\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{n_2^2}\right]$
  
  For 1st $1^{\text{st }}$ line n2=3${\mathrm{n}}_2=3$
  
  so, 1λ=R[122−132]$\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$
  
  for series limit n2=∞${\mathrm{n}}_2=\mathrm{\infty }$
  
  so, 1λ=R4$\frac{1}{\lambda }=\frac{\mathrm{R}}{4}$
  
  →$\to$ It lies in visible region
  
  
  

• Paschen series:

  If n1=3&n2=4,5,6,…$n_1=3\mathrm{\&}{n}_2=4,5,6,\dots$
  
  So, 1/λ=R[132−1n22]$1/\lambda =\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{n_2^2}\right]$
  
  For 1st $1^{\text{st }}$ line n2=4${\mathrm{n}}_2=4$
  
  so, 1λ=R[132−142]$\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{4^2}\right]$
  
  for series limit n2=∞${\mathrm{n}}_2=\mathrm{\infty }$
  
  so, 1λ=R9$\frac{1}{\lambda }=\frac{\mathrm{R}}{9}$
  
  →$\to$ It lies in IR region
  
  
  

• Brackett series:

  If n1=4&n2=5,6,7,…$n_1=4\mathrm{\&}{n}_2=5,6,7,\dots$
  
  So, 1/λ=R[142−1n22]$1/\lambda =\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{n_2^2}\right]$
  
  For 1st $1^{\text{st }}$ line n2=5${\mathrm{n}}_2=5$
  
  so, 1λ=R[142−152]$\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{5^2}\right]$
  
  for series limit n2=∞${\mathrm{n}}_2=\mathrm{\infty }$
  
  so, 1λ=R16$\frac{1}{\lambda }=\frac{\mathrm{R}}{16}$
  
  →$\to$ It lies in IR region
  
  

Excitation Energy/ Potential

Amount of energy required to jump the electrons from ground state to higher energy state is known as excitation energy. Required potential to do so is called excitation potential.

n=1$n=1$ to n=2$n=2$

Example:

1st$1^{st}$ excitation energy is

E=E2−E1=−3.4+13.6=10.2ev$$\begin{gathered} E=E_2-E_1=-3.4+13.6 \\ =10.2ev \end{gathered}$$

Excitation potential

Ve =10.2volt$V_e=10.2\text{volt}$

Ionization Energy/Potential

Amount of energy required to jump the electrons from any energy state to infinite energy state is known as inonization energy. Required potential to knock out the electrous from an atom is known as ionization potential.

When an e−$e^{-}$ ionized from n=1$n=1$

Requried ionization energy is:

(Ei)=E∞−E1=0−(−13.6)=13.6ev$\begin{array}{rl}(E_i)& =E_{\mathrm{\infty }}-E_1\\ & =0-(-13.6)\\ & =13.6ev\end{array}$

∴$\therefore$ ionization potential (Vi)=13.6volt$(V_i)=13.6\text{volt}$

And so on.

Heisenberg's uncentainty principle

Heisenberg's uncentainty principle states that the product of uncertainity of two conjugate variables is always greater than or equal to h/2π$h/2\pi$.

i.e. Δx×Δp≥h/2π$\mathrm{\Delta }\mathrm{x}\times \mathrm{\Delta }\mathrm{p}\ge \mathrm{h}/2\pi$

x=$\mathrm{x}=$ position
Δx=$\mathrm{\Delta }\mathrm{x}=$ uncertainity is position
P=$\mathrm{P}=$ Momentum
ΔP=$\mathrm{\Delta }\mathrm{P}=$ uncentainity is momentum

Similary for energy & time
ie. ΔE×Δt≥h/2π$\mathrm{\Delta }\mathrm{E}\times \mathrm{\Delta }\mathrm{t}\ge \mathrm{h}/2\pi$

De-Broglie wave Equation

• He explained dual nature of matter (wave nature of matter)
  
  
• If matter posses wave nature, it will be associated particular wavelength ie λ=h/p$\lambda =h/p$
  
  h = planck's constant
  p = momentum
  
  Let,
  m = mass of matter
  V = velocity of matter
  
  Therefore, de Broglie wavelength for matter is:
  λ=h/mv$\lambda =h/mv$
  
  
• When an e−$e^{-}$ is accelerated through p.d. of V then de-Broglie's wavelength for an electron is:
  
  P.E. gain by e−=eV$e^{-}=eV$
  
  Accelerated K.E.=12mv2$K.E.=\frac{1}{2}m{v}^2$
  
  We have P.E = K.E
  
  eV=1/2mv2$\mathrm{e}\mathrm{V}=1/2{\mathrm{m}\mathrm{v}}^2$
  ⇒v=2eVm−−−−√$\Rightarrow \mathrm{v}=\sqrt{\frac{2eV}{m}}$
  
  We have, de-Broglie wave length for an electron is:
  
  λele=hmv=hm2eVm−−−−√${\lambda }_{ele}=\frac{h}{mv}=\frac{h}{m\sqrt{\frac{2eV}{m}}}$
  
  =h2meV−−−−−√=h2mEk−−−−−√$=\frac{h}{\sqrt{2meV}}=\frac{h}{\sqrt{2m{E}_k}}$
  
  =12.27v√A˚