Atomic Structure

Atomic Structure Notes

Introduction to Atom

• Classical mechanics were inadequate for atomic and molecular systems.

• Quantum theory, proposed by Max Planck and applied by Einstein and Bohr, is crucial for the modern atomic concept.

• Black body radiation, photoelectric effect, and hydrogen atomic spectra are explained by Planck, Einstein, and Bohr's theories.

• These theories posit interactions between matter and radiation occur in definite energy packets called quanta.

• Wave mechanics evolved from this, showing the dual nature of matter and energy.

Dalton's Atomic Theory

• Based on the law of mass conservation and definite proportions.

• Salient features:

  • Elements are composed of small particles called atoms.

  • Atoms of an element are alike but differ from other elements.

  • Atoms are ultimate particles with characteristic mass but are structureless.

  • Atoms are indestructible.

  • Atoms take part in chemical reactions to form molecules.

Thomson's Model of Atom (1904)

• Thomson proposed the first detailed atomic model.

• An atom consists of a uniform sphere of positive charge with electrons present at some places.

• Known as the 'Plum-Pudding model'.

• Drawbacks:

  • Mass is considered evenly spread over the atom.

  • It is a static model, not reflecting electron movement.

Rutherford's α-Scattering Experiment

• Observations:

  • Most α-particles (99.9%) went straight without deflection.

  • A few were deflected through small angles.

  • Very few (1 in 20,000) suffered large deflections (more than 90°) or came back (180° deflection).

• Conclusions:

  • Atoms have large empty spaces.

  • Positive charge is concentrated in a small volume called the nucleus.

  • The nucleus is rigid, causing α-particles to recoil upon direct collision.

• Applications:

  • Nuclear atomic model: a heavy, positively charged nucleus with protons.

  • Nucleus volume is a tiny fraction of the atom's total volume.

  • Empty space around the nucleus with electrons.

  • Number of electrons equals the number of protons.

  • Extra nuclear part is responsible for the volume.

  • Electrons revolve around the nucleus in closed orbits at high speeds, similar to a solar system.

• Radius Relationships:

  • $r_{atom} ≈ 10^{-8} cm$

  • $r_{nucleus} ≈ 10^{-13} cm$

  • $r<em>{atom} ≈ 10^5 r</em>{nucleus}$

  • $R ∝ A^{1/3}$

  • $R = R_0 A^{1/3} cm$

  • $R_0 = 1.33 × 10^{-13}$

  • $A$ = mass number ($p + n$)

• Volume Relationships:

  • $\frac{vol. \, of \, the \, atom}{vol. \, of \, the \, nucleus} = \frac{\frac{4}{3}πr<em>A^3}{\frac{4}{3}πr</em>N^3} = \frac{(10^{-8})^3}{(10^{-13})^3} = 10^{15}$

Distance of Closest Approach

• α-particle slows down as it approaches the nucleus, converting kinetic energy into electrostatic potential energy.

• The distance of closest approach ($r_0$) occurs when the α-particle comes to rest.

• Potential energy at point B: $P.E. = \frac{1}{4πε<em>0} \frac{(2e)(Ze)}{r</em>0}$

• Kinetic energy of α-particle: $K.E. = \frac{1}{2} m v^2$

• Equating both, $r<em>0 = \frac{2Ze^2}{4πε</em>0 × K.E.}$

• Example: For an α-particle with K.E. = 7.7 MeV scattered by gold (Z = 79), $r_0 ≈ 3 × 10^{-14} m$.

• This indicates nuclear dimensions cannot be greater than $3 × 10^{-14} m$.

• Drawbacks of Rutherford Model:

  • Could not explain the stability of an atom.

  • According to Maxwell, electrons should lose energy continuously and fall into the nucleus, making the atom unstable.

  • Observed spectrum should be continuous, but it is discontinuous with definite frequencies.

Wave Theory

• A wave is a periodic disturbance in space or a medium, involving elastic displacement or changes in physical quantities.

• Wave motion represents the propagation of a periodic disturbance carrying energy.

• The wave travels at right angles to the vibratory motion of the object.

Wave Characteristics

• Crest: Uppermost point of the wave.

• Trough: Lowermost point of the wave.

• Wavelength ($\lambda$):

  • Distance between two nearest crests or troughs.

  • Measured in Å, pm, nm, cm, m.

  • $1 Å = 10^{-10} m, 1 pm = 10^{-12} m, 1 nm = 10^{-9} m, 1 cm = 10^{-2} m$

• Frequency ($ν$):

  • Number of waves passing through a point in 1 second.

  • Measured in Hertz (Hz), $sec^{-1}$, or cycles per second (cps).

  • $1 Hertz = 1 sec^{-1} = 1 cps$

• Time Period ($T$):

  • Time taken by a wave to pass through one point.

  • $T = \frac{1}{ν} sec$

• Velocity ($c$):

  • Distance covered by a wave in 1 second.

  • $c = \frac{λ}{T} = λν$

  • Since $c$ is constant, $ν = \frac{c}{λ}$, implying $ν ∝ \frac{1}{λ}$.

• Wave Number ($\overline{ν}$):

  • Reciprocal of the wavelength, indicating the number of waves in 1 cm.

  • $\overline{ν} = \frac{1}{λ}$

  • Measured in $cm^{-1}, m^{-1}$.

  • $1 \, cm^{-1} = 100 \, m^{-1}$

  • $ν = \frac{c}{λ} = c\overline{ν}$

• Amplitude ($a$):

  • Height of the crest or depth of the trough.

Electromagnetic Waves (EM waves) or Radiant Energy

• Energy transmitted from one body to another in the form of waves, traveling at the speed of light ($3 × 10^8 m/s$).

• Do not need a medium for propagation.

• Examples: Radio waves, microwaves, infrared rays, visible rays, ultraviolet rays, x-rays, gamma rays, and cosmic rays.

Illustrations

• Illustration 2: Calculating the wavelength of an EM wave emitted by All India Radio.

  • Given frequency ($ν$) = 1368 kHz = $1368 × 10^3 Hz$

  • Using $λ = \frac{C}{ν}$, where $C = 3 × 10^8 m/sec$

  • $λ = \frac{3 × 10^8}{1368 × 10^3} = 219.3 \, m$

• Illustration 3: Calculating wave number and frequency of yellow radiations with wavelength of 5800 Å.

  • $\overline{ν} = \frac{1}{λ} = \frac{1}{5800 × 10^{-8}} = 17241.4 \, cm^{-1}$

  • $v = c\overline{ν} = 3 × 10^{10} × 1.7 × 10^4 = 5.1 × 10^{14} \, sec^{-1}$

• Illustration 4: Calculating time for a radio wave to travel from Mars to Earth.

  • Distance = $8 × 10^7 km = 8 × 10^{10} m$

  • Time = $\frac{Distance}{Velocity} = \frac{8 × 10^{10}}{3 × 10^8} = 266 \, sec$

• Illustration 5: Calculating frequency of a photon with wavelength 2225 Å.

  • $v = \frac{Velocity}{Wavelength} = \frac{3 × 10^8}{2225 × 10^{-10}} = 1.349 × 10^{15} \, sec^{-1}$

Planck's Quantum Theory

• Electromagnetic theory regarded radiant energy as a continuous flow of energy in waves but failed to explain:

  • Black body radiations.

  • Photoelectric effect.

  • Line spectrum emitted by excited atomic gases.

• Features of Planck's quantum theory:

  • Radiant energy is emitted or absorbed discontinuously in small, discrete packets called quanta.

  • For light, the smallest packet is a 'photon'.

  • Energy of each quantum is directly proportional to the frequency of the radiation: E ∝ ν => E = hν = \frac{hc}{λ} = hc\overline{ν}

    • $h$ is Planck's constant.

  • $h = 6.626 × 10^{-34} J \, sec = 6.626 × 10^{-27} erg \, sec$

  • Total energy transmitted is an integral multiple of the energy of a quantum: $E = nhν$, where $n$ is an integer.

Illustrations (Planck's Quantum Theory)

• Illustration 6: Calculating the energy of a photon of sodium light.

  • Given $λ = 5.862 × 10^{-16} m$

  • $E = \frac{hc}{λ} = \frac{6.6 × 10^{-34} × 3 × 10^8}{5.862 × 10^{-16}} = 3.38 × 10^{-10} \, Joules$

• Illustration 7: Calculating the frequency and energy of a photon with a wavelength of 4000 Å.

  • $ν = \frac{c}{λ} = \frac{3 × 10^8}{4 × 10^{-7}} = 7.5 × 10^{14} \, sec^{-1}$

  • $E = hν = 6.626 × 10^{-34} × 7.5 × 10^{14} = 4.96 × 10^{-19} \, Joule$

• Illustration 8: Calculating wavelength and frequency of a photon with an energy of 2 eV.

  • $E = 2 \, eV = 3.204 × 10^{-19} J$

  • $λ = \frac{hc}{E} = \frac{6.626 × 10^{-34} × 3 × 10^8}{3.204 × 10^{-19}} = 6.204 × 10^{-7} \, m$

  • $ν = \frac{c}{λ} = \frac{3 × 10^8}{6.204 × 10^{-7}} = 4.8 × 10^{14} \, sec^{-1}$

• Illustration 9: Comparing the energy of a violet light photon and a red light photon.

  • $E_{violet} = \frac{hc}{4000 × 10^{-10}} = 4.97 × 10^{-19} \, Joule$

  • $E_{red} = \frac{hc}{7000 × 10^{-10}} = 2.8 × 10^{-19} \, Joule$

  • E{violet} > E{red}

• Illustration 10: Calculating the number of photons needed to provide 1 Joule of energy.

  • $n = \frac{Eλ}{hc} = \frac{1 × 5000 × 10^{-10}}{6.626 × 10^{-34} × 3 × 10^8} = 2.5 × 10^{18} \, photons$

Photoelectric Effect

• Discovered by J.J. Thomson.

• Electrons are ejected from a metal surface when light of a certain frequency strikes it; these electrons are called photoelectrons.

• Metals with low ionization energy (e.g., Caesium) show this effect under visible light, while others require ultraviolet light.

• Experimental Findings:

  • Electrons come out as soon as the light strikes the metal surface.

  • There is a minimum (threshold) frequency ($ν_0$) required for ejection, which varies with the metal's nature.

  • Higher frequency light yields more energetic photoelectrons.

  • Increased intensity of light increases photoelectric current, but electron energies remain the same.

• Light has energy particles (quanta) with energy $hν$.

  • If ν < ν_0, no ejection occurs.

  • If ν > ν_0, the excess energy becomes kinetic energy (K.E.) of the electron.

  • $hν = hν_0 + K.E.$

  • $K.E.<em>{max} = hν - hν</em>0$

• $hν_0$: Work function ($\phi$), which is constant for a particular metal.

• The kinetic energy of photoelectrons increases linearly with the frequency of incident light.

• Stopping Potential ($V_0$):

  • The minimum reverse potential applied to stop photocurrent.

  • $K.E.<em>{max} = eV</em>0 = hν - \phi$

Illustrations (Photoelectric Effect)

• Illustration 11: Calculating energy, kinetic energy, and velocity of photoelectrons.

  • Given: $λ = 3000 Å$, Work function ([\phi]) = 2.20 eV

  • Energy of the photon: $E = \frac{hc}{λ} = \frac{6.6 × 10^{-34} × 3 × 10^8}{3 × 10^{-7}} = 6.6 × 10^{-19} J = 4.125 \, eV$

  • Kinetic energy of emitted photoelectron: $K.E. = 4.125 - 2.20 = 1.925 \, eV = 3.08 × 10^{-19} J$

  • Velocity of the photoelectron: $v = \sqrt{\frac{2 × K.E.}{m}} = \sqrt{\frac{2 × 3.08 × 10^{-19}}{9.1 × 10^{-31}}} = 8.22 × 10^5 \, ms^{-1}$

• Illustration 12: Calculating the de Broglie wavelength of electrons.

  • Given: $λ = 2000 Å$, Threshold = $4000 Å$

  • $K.E. = \frac{hc}{2000 × 10^{-10}} - \frac{hc}{4000 × 10^{-10}} = 4.969 × 10^{-19} \, Joule$

  • $λ = \frac{h}{mv} = \frac{6.626 × 10^{-34}}{9.51 × 10^{-25}} = 0.696 × 10^{-9} \, m$

Bohr's Atomic Model

• Important Formulae:

  • Coulombic force: $F<em>e = \frac{kq</em>1q_2}{r^2}$

  • Centrifugal force: $F_c = \frac{mv^2}{r}$

  • Angular momentum: $L = mvr$

• Based on quantum theory and classical physics.

• Postulates:

  • Atom has a nucleus with protons and neutrons at the center.

  • Electrons revolve around the nucleus like planets around the sun in circular paths.

  • Electrostatic force equals centrifugal force: $\frac{KZe^2}{r^2} = \frac{mv^2}{r}$

  • Angular momentum is quantized: $mvr = \frac{nh}{2π}$, where n = 1, 2, 3, …

  • Orbits where electrons can revolve are stationary with constant energy.

  • Emission or absorption of energy occurs when electrons jump between stationary states: $\Delta E = E<em>{final \, state} - E</em>{initial \, state}$

  • Energy is absorbed when electrons jump from inner to outer orbits and emitted vice versa.

Radii of Various Orbits of Hydrogen Atom

• Electron mass = ‘m’, charge = ‘e’ revolving around nucleus with charge Ze with tangential velocity v.

• Electrostatic force: $F = K\frac{Ze^2}{r^2}$ where $K = \frac{1}{4πε_0} = 9 × 10^9 Nm^2C^{-2}$

• Centrifugal force: $F = \frac{mv^2}{r}$

• For a stable orbit, $K\frac{Ze^2}{r^2} = \frac{mv^2}{r}$

• From Bohr's postulate, mvr = \frac{nh}{2π} => v = \frac{nh}{2πmr}

• Radius: $r = \frac{n^2h^2}{4π^2mKZe^2}$, where $n = 1, 2, 3, …$

• Radius of the smallest orbit for hydrogen ($n = 1, Z = 1$): $r<em>0 = 0.529 Å$ Radius of nth orbit: $r</em>n = 0.529 × \frac{n^2}{Z} Å$

Calculation of Energy of an Electron

• Total Energy (E) is the sum of Kinetic Energy (K.E) and Potential Energy (P.E.).

• Kinetic Energy: $K.E = \frac{1}{2}mv^2 = \frac{KZe^2}{2r}$

• Potential Energy = $-\frac{KZe^2}{r}$

• Total Energy (E) = $-\frac{KZe^2}{2r}$

• $E = -21.8 × 10^{-12} × \frac{Z^2}{n^2} erg \, per \, atom$

• $E = -21.8 × 10^{-19} × \frac{Z^2}{n^2} J \, per \, atom$

• $E = -13.6 × \frac{Z^2}{n^2} eV \, per \, atom$ (where $1 eV = 1.602 × 10^{-19} J$)

• $E = -313.6 × \frac{Z^2}{n^2} Kcal/mole$ (where $1 cal = 4.18 J$)

• A negative energy indicates electron in the atom has less energy than a free electron (at infinite distance from the nucleus).
  When n = ∞, E = 0, the atom is ionized.

Calculation of Velocity

• Based on $mvr = \frac{nh}{2π}$, we get $v = \frac{nh}{2πmr}$

• Velocity $v = 2.18 × 10^8 \frac{Z}{n} cm/sec$

Terms Associated with Bohr's Model

• Ground State: Lowest energy state of an atom or ion.

  • Ground state of H-atom = -13.6 eV.

  • Ground state of $He^+$ ion = -54.4 eV.

• Excited State: States of atom other than the ground state.

  • n = 2: first excited state.

  • n = 3: second excited state.

  • n = n+1: nth excited state.

• Ionization Energy (IE): Minimum energy required to move an electron from ground state to n = ∞.

  • IE of H-atom = 13.6 eV.

  • IE of $He^+$ ion = 54.4 eV.

  • IE of $Li^{+2}$ ion = 122.4 eV.

• Ionization Potential (IP): Potential difference to accelerate a free electron to have kinetic energy = ionization energy.

  • I.P. of H atom = 13.6 V.

  • I.P. of $He^+$ Ion = 54.4 V.

• Excitation Energy: Energy to move an electron from ground state to any other state of the atom.

  • Excitation energy of 2nd state = 10.2 eV.

• Excitation Potential: Potential difference to accelerate an electron to have kinetic energy = excitation energy.

  • Excitation potential of 3rd state = 12.09 V.

• Binding Energy or Separation Energy: Energy required to move an electron from any state to n = ∞.

  • Binding energy of ground state = I.E. of atom or Ion.

Illustrations (Bohr Model)

• Illustration 13: Calculating radii of Bohr's orbits for hydrogen.

  • $r = 0.529 × \frac{n^2}{Z}$

  • (a) Radius of 1st orbit: $r = 0.529 Å$

  • (b) Radius of 2nd orbit: $r = 2.116 Å$

  • (c) Radius of 3rd orbit: $r = 4.761 Å$

  • (d) Radius of 4th orbit: $r = 8.464 Å$

• Illustration 14: Calculating the radius ratio of 2nd orbit of hydrogen and 3rd orbit of $Li^{+2}$.

  • H: $r_2 = 0.529 × \frac{2^2}{1}$

  • $Li^{+2}$: $r_3 = 0.529 × \frac{3^2}{3}$

  • Ratio = 4:3

• Illustration 15: Radius ratio of $Li^{+2}$ is 1:9 indicating which orbits ?

  • Ratio is $1^2 : 3^2$ so K & M

• Illustration 16: Calculating the energy of $Li^{+2}$ atom for 2nd excited state.

  • $E = -13.6 × \frac{Z^2}{n^2} = -13.6 × \frac{3^2}{3^2} = -13.6 \, eV$

• Illustration 17: Calculating the ratio of energies of $He^+$ for 1st & 2nd excited state.

• Ratio is $\frac{(2)^2}{(2)^2} : \frac{(2)^2}{(3)^2} = 1: \frac{4}{9}$

• Illustration 18: Calculating the required energy to excite H from the ground state to the 1st excited state.

  • $E<em>2 - E</em>1 = -3.4 + 13.6 = 10.2 \, eV$

• Illustration 19: Finding K.E, P.E, orbit radius and velocity of electron in H atom with total energy of -1.51 eV.

  • K.E = 1.51 eV.

  • P.E = -3.02 eV.

  • Orbit = 3rd

  • r = 4.761 Å

  • v = $0.729 × 10^8 cm/sec$

• Illustration 20: Calculating the velocity of an electron in the 3rd orbit of the ion.$Li^{+2}$ Also calculate the number of revolutions per second that it makes around the nucleus.

• No. of revolutions/sec = $0.2187 × 10{16} rev/sec$

Spectrum

• Electromagnetic Spectrum (EM Spectrum)

  • Arrangement of various EM waves by increasing frequency or decreasing wavelength.

  • low v low E Radio Waves, Micro waves, IR, Visible, UV, X-ray, Gamma, Cosmic rays high v high E

• Spectrum

  • Pattern (photograph) obtained when radiation is passed through a spectroscope (prism) for dispersion.

• Classification of Spectrum

  • Emission (Continuous, line, band)

  • Absorption (line, band)

Emission Spectrum

• Radiation emitted from incandescence source passed through a prism and received on the screen.

• Emission Continuous Spectrum:
  Narrow beam of white light through a prism dispersed into 7 colors (violet to red).

• Emission Line Spectrum
  Atomic gas raised to incandescence source emits radiation examined through a spectroscope a spectrum is obtained which have well defined lines for a definite wavelength with dark space: Emission line spectrum.

• Emission Band Spectrum
  Molecular gas absorbs energy for electron transition, rotational, vibrational and translational then emits radiations examined through a spectroscope a spectrum is obtained on the screen on the screen, which are group of closely packed lines called Bands, therefore this type of Emission spectrum is called as emission band spectrum. separated by dark space: Emission Line spectrum.

Absorption Spectrum

• White light first passed through a chemical substance or gas and then analysed by spectroscope.

• Some dark lines are obtained in otherwise continuous spectrum: Absorption Spectrum
  If white light is passed through atomic gas then: Absorption line spectrum
  If white light is passed through molecular gas then: Absorption band spectrum.

Hydrogen Line Spectrum

• Electric excitation applied to atomic hydrogen gas at low pressure emits bluish light.

• Light passed through a prism gives a spectrum of isolated sharp lines.

• Lines lie in Visible, Ultraviolet and Infra-red regions, grouped into different series.

• KEY POINTS:

  • First line/Starting line/Initial line (max. \[ λ ] and \[ ν ] min )

  • Last line/limiting line/Series limit (λ min and \[ ν ]max.)

  • First line of any series = α line

  • Second line of any series = β line

  • Third line of any series = y line
    Total no. of emission lines between n₂ & n₁ (n₂ > n₁) = $\frac{(n<em>2 - n</em>1)(n<em>2-n</em>1 +1)}{2}$
    For transition from any orbit 'n' to n = 1, total no. of emission lines = $\frac{n(n - 1)}{2}$

Formulae

• \overline{ν} = \frac{1}{λ} = RH \left( \frac{1}{n1^2} - \frac{1}{n2^2} \right) × Z^2 value of RH =109677 cm⁻¹= $10967700m^{-1}$
  =$\frac{109700}{cm^{-1}}$
  λ = 912Å/Z2

Illustrations Hydrogen spectrum

Example 21. Electron moves from 7 to 1 by multi - steps total number of lines in spectra
Number of transition for each level:
Lyman=(n₂ -1 ) = 7 - 1 = 6;
Balmer =(n₂ -2 ) = 7 - 2 = 5;
Paschen =(n₂ -3 ) = 7 - 3 = 4;
Bracket =(n₂ -4 ) = 7 - 4 = 3;
Pfund=(n₂ -5 ) = 7 - 5 = 2
Humphrey=(n₂ -6 ) = 7 - 6 = 1, Total no. of transition = (6+5+4+3+2+1) =21

• Illustration 22. e⁻ moves from 6th to 3rd . Total lines, UV lines, Visible lines, IR lines in spectrum:
  Total = (6 transition), For UV = zero, Visible: zero,IR=(3+2+1)=6;
  In Balmer series 3rd line 5th to 2nd, Total = 6 line

• Illustration 23. If the spectrum of a Hydrogen if e- moves from n orbit to the 1 .If total numbers of lines is10. Find n. Number of lines in H -atom:
  $\frac{n(n-1)}{2}=10 \Rightarrow n=5$

• Illustration 2. Calculate the wavelength of 1st line of Balmer series in Hydrogen spectrum.
  ν =R[1/4-1/9]; solving λ = 6566.4 Å
  What will be frequency of last line of the Lyman series of hydrogen spectrum ?v=3.29 1015 sec⁻¹

de Broglie

• French physicist, Louis de Broglie suggested that if the nature of light is both that of a particle and of a wave, then this dual behavior should be true for the matter also.
  mv
  h Planck constant/ momentum of electron h√2m(K.E), Decreasing wave length lighter particle have a wave length than heavier V potential= $λ = h/√√2mQV$
  The circumference of any nth the 2 r equal $nλ$ electron orbit.
  Illustration.
  If X=40,calculate the wave length λ Y= 2x40 x=80
  Calculate the de Broglie wavelength, of a ball mass 0.1 moving with the speed of 30/s λ=$\frac{6.6x10^{-34}}{0.1 x 30}=2.2 x 10^{-34}$m
  The wave length if 9.1 x1-31 velocity tenth light λ =2.426

Heisenberg Uncertainty Principle

It is impossible to measures simultaneously the exact position with the exact momentum of the small body has an electron. The uncertainty to measure velocity/momentum $Ax Ap \geq \frac{h}{4π}$
If Ax = 0 Av= ∞ the value is uncertainty the product is neglectable.In energy with time is written 4E 4t>h/4π
Example,Diameter nuclei of atom, 10-15.Max, uncertainty, electron =10-15 , so speed, 5 , 8.10/s higher with speed light.
The uncertainty .0 is mass 99.90 so Av =
Why electron are not inside. Atomic nuclei of the is maximum umber.
If electron as the mass m uncertainty product is replaced concept the replacing the concept of concept of orbit product by product product by product probability .
What is the uncertainty in the electron 9.10090 and value is 6.253 A = velocity 0.5 82.88 10.30
. Ax Ap>A/P same will same.

Wave mechanical model atom

Orbital, region space probability by an electron and atom high around. different draw the region of 1
The used is Schrö nursing. Schödinger equation play laws of in law of energy, it predict future system number give system the distribution ,,, at gives all measurable calculation expectation.
Schrödinger, = three equation is continuous find the number of solutions it it.

Quantum number

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Radial probability

likelihood of the from gives can shell distribution. function.
It the between distribution. has the volume the radial.
Electrons density has node points number

Angular probability

Angular shape is orientation.
Number of shells, number each with number
Describe subshell number is

RULES FOR FILLING

Aufbau Principle building up added orbital in order lowest

Pauli's Exclusion

No equal electron two orbital that is more spins

Hund's Maximum Multiplicity

E distributed the manner unpaired election
Configuration different

Exception of Aufbau principle greater

The stability the filled and electronic configurations electrons their the of of or not for positive a by adding loss from

Properties Paramagnetism

Attracted and that that that retain retain and and it zero. Ferromanism strong the retain cannot states.

• Total spin n unpaired.
  total that μ effective = √ n (n 4 s + 1) is and

I am unable to generate graphs, but here's a structured approach to remember formulas and relationships within atomic structure, categorized for clarity:

1. Wave Characteristics:

   • $c = λν$ (Velocity, Wavelength, Frequency)

   • $ν = \frac{c}{λ}$ (Frequency in terms of speed of light and wavelength)

   • $\overline{ν} = \frac{1}{λ}$ (Wave Number)

2. Planck's Quantum Theory:

   • $E = hν$ (Energy of a quantum/photon)

   • $E = \frac{hc}{λ}$ (Energy in terms of wavelength)

   • $E = hc\overline{ν}$ (Energy using wave number)

3. Photoelectric Effect:

   • $hν = hν_0 + K.E.$ (Energy of photon = Work function + Kinetic Energy)

   • $K.E._{max} = hν - \phi$ (Work function is \phi)

   • $K.E._{max} = eV<em>0$ (Stopping Potential, $V</em>0$)

4. Rutherford's Model & Related Radii:

   • $R \propto A^{1/3}$ (Nuclear radius related to mass number)

   • $R = R<em>0 A^{1/3} cm$ (where $R</em>0 = 1.33 × 10^{-13} cm$)

5. Bohr's Atomic Model:

   • $\frac{KZe^2}{r^2} = \frac{mv^2}{r}$ (Electrostatic force equals centrifugal force)

   • $mvr = \frac{nh}{2π}$ (Quantization of angular momentum)

   • $r = 0.529 × \frac{n^2}{Z} Å$ (Radius of nth orbit)

   • $E = -13.6 × \frac{Z^2}{n^2} eV$ (Energy of an electron in nth orbit)

   • $v = 2.18 × 10^8 \frac{Z}{n} cm/sec$ (Velocity of electron)

6. Hydrogen Spectrum

   • $\overline{ν} = \frac{1}{λ} = R<em>H \left( \frac{1}{n</em>1^2} - \frac{1}{n_2^2} \right) × Z^2$

   • Total no. of emission lines between $n<em>2$ & $n</em>1$ (n2 > n1) = $\frac{(n<em>2 - n</em>1)(n<em>2-n</em>1 +1)}{2}$

   • For transition from any orbit 'n' to n = 1, total no. of emission lines = $\frac{n(n - 1)}{2}$

7. de Broglie Wavelength:

   • $λ = \frac{h}{mv}$

8. Heisenberg Uncertainty Principle:

   • $\Delta x \Delta p \geq \frac{h}{4π}$

Remember to understand what each variable represents and how the formulas relate to the concepts they describe. Focus on how changing one variable affects others within each formula. Regular practice and application will solidify these relationships in your memory.