Example 1: Electron Velocity in Thomson's Crossed Fields

In Thomson's cathode ray experiment, an electron moves through perpendicular electric and magnetic fields of strengths 2×1052 \times 10^5 V/m and 0.01 T respectively. When the forces balance, calculate the velocity of the electron.

Solution: For no deflection, eE=evBeE = evB v=EB=2×1050.01=2×107 m/sv = \frac{E}{B} = \frac{2 \times 10^5}{0.01} = 2 \times 10^7 \text{ m/s}

Final Answer: 2×1072 \times 10^7 m/s

Example 2: Charge on Electron

Using the known charge-to-mass ratio of the electron, em=1.76×1011\frac{e}{m} = 1.76 \times 10^{11} C/kg, and electron mass 9.109×10−319.109 \times 10^{-31} kg, calculate the charge on the electron.

Solution: e=em×m=1.76×1011×9.109×10−31e = \frac{e}{m} \times m = 1.76 \times 10^{11} \times 9.109 \times 10^{-31} e=1.603×10−19 Ce = 1.603 \times 10^{-19} \text{ C}

Final Answer: 1.6×10−191.6 \times 10^{-19} C

Example 3: Closest Approach in Rutherford's Experiment

In Rutherford's alpha-particle scattering experiment, an alpha particle with kinetic energy 5 MeV approaches a gold nucleus head-on. Calculate the distance of closest approach.

Solution: For head-on collision, KE=14πϵ0Z1Z2e2rmin⁡KE = \frac{1}{4\pi\epsilon_0}\frac{Z_1 Z_2 e^2}{r_{\min}} Here, Z1=2Z_1 = 2, Z2=79Z_2 = 79, KE=5×1.6×10−13KE = 5 \times 1.6 \times 10^{-13} J.

rmin⁡=8.99×109×2×79×(1.6×10−19)28.0×10−13r_{\min} = \frac{8.99 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{8.0 \times 10^{-13}} rmin⁡=4.54×10−14 mr_{\min} = 4.54 \times 10^{-14} \text{ m}

Final Answer: 4.54×10−144.54 \times 10^{-14} m = 45.4 fm

Example 4: Nuclear Charge from Rutherford Scattering

An alpha particle of kinetic energy 5 MeV approaches a nucleus with impact parameter 2 fm and is deflected by 60°. Using Rutherford scattering, determine the nuclear charge number ZZ.

Solution: cot⁡(θ2)=2KE bkZ1Z2e2\cot\left(\frac{\theta}{2}\right) = \frac{2KE\,b}{k Z_1 Z_2 e^2} For θ=60∘\theta = 60^\circ, cot⁡30∘=3\cot 30^\circ = \sqrt{3}.

Rearranging, Z2=2KE bkZ1e2cot⁡(θ/2)Z_2 = \frac{2KE\,b}{k Z_1 e^2 \cot(\theta/2)} Substituting KE=5×106×1.6×10−19KE = 5 \times 10^6 \times 1.6 \times 10^{-19} J, b=2×10−15b = 2 \times 10^{-15} m, Z1=2Z_1 = 2: Z2≈79Z_2 \approx 79

Final Answer: Z=79Z = 79 (gold nucleus)

Example 5: Atomic Number and Mass Number

A nucleus has 25 protons and 30 neutrons. Write the symbol notation, identify the element, and calculate the mass defect.

Solution: Atomic number Z=25Z = 25, mass number A=25+30=55A = 25 + 30 = 55. Element with Z=25Z=25 is Mn.

Symbol notation: 2555Mn^{55}_{25}\text{Mn}

Calculated mass: (25×1.0073)+(30×1.0087)=55.4435 u(25 \times 1.0073) + (30 \times 1.0087) = 55.4435 \text{ u} Actual mass = 54.9380 u Δm=55.4435−54.9380=0.5055 u\Delta m = 55.4435 - 54.9380 = 0.5055 \text{ u} Binding energy: 0.5055×931.5=470.8 MeV0.5055 \times 931.5 = 470.8 \text{ MeV}

Final Answer: 2555Mn^{55}_{25}\text{Mn}; Manganese-55; Mass defect = 0.5055 u; Binding energy = 470.8 MeV

Example 6: Isotopes, Isobars, and Isotones

Consider the following atoms: 1632S^{32}_{16}\text{S}, 1532P^{32}_{15}\text{P}, 2040Ca^{40}_{20}\text{Ca}, 1940K^{40}_{19}\text{K}, 1735Cl^{35}_{17}\text{Cl}, 1737Cl^{37}_{17}\text{Cl}. Classify them as isotopes, isobars, or isotones.

Solution: Isotopes: same ZZ, different AA:

  • 1735Cl^{35}_{17}\text{Cl} and 1737Cl^{37}_{17}\text{Cl}

Isobars: same AA, different ZZ:

  • 1632S^{32}_{16}\text{S} and 1532P^{32}_{15}\text{P}
  • 2040Ca^{40}_{20}\text{Ca} and 1940K^{40}_{19}\text{K}

Isotones: same neutrons NN:

  • 2040Ca^{40}_{20}\text{Ca} has N=20N=20
  • 1737Cl^{37}_{17}\text{Cl} has N=20N=20

Final Answer: Isotopes: Cl-35 and Cl-37; Isobars: (S-32, P-32) and (Ca-40, K-40); Isotones: Ca-40 and Cl-37

Example 7: Wavelength, Frequency, Wavenumber, and Energy

A photon has a wavelength of 500 nm. Calculate its frequency, wavenumber, and energy.

Solution: λ=500×10−9=5×10−7 m\lambda = 500 \times 10^{-9} = 5 \times 10^{-7} \text{ m} ν=cλ=3×1085×10−7=6×1014 Hz\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{5 \times 10^{-7}} = 6 \times 10^{14} \text{ Hz} νˉ=1λ=2×106 m−1\bar{\nu} = \frac{1}{\lambda} = 2 \times 10^6 \text{ m}^{-1} E=hν=6.626×10−34×6×1014=3.98×10−19 JE = h\nu = 6.626 \times 10^{-34} \times 6 \times 10^{14} = 3.98 \times 10^{-19} \text{ J} In eV: E=3.98×10−191.6×10−19=2.49 eVE = \frac{3.98 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.49 \text{ eV}

Final Answer: Frequency = 6×10146 \times 10^{14} Hz; Wavenumber = 2×1062 \times 10^6 m−1^{-1}; Energy = 3.98×10−193.98 \times 10^{-19} J = 2.49 eV

Example 8: Photoelectric Effect

A metal surface has a threshold frequency of 8×10148 \times 10^{14} Hz. When light of frequency 1.5×10151.5 \times 10^{15} Hz falls on it, what is the stopping potential?

Solution: W=hν0=6.626×10−34×8×1014=5.30×10−19 JW = h\nu_0 = 6.626 \times 10^{-34} \times 8 \times 10^{14} = 5.30 \times 10^{-19} \text{ J} E=hν=6.626×10−34×1.5×1015=9.94×10−19 JE = h\nu = 6.626 \times 10^{-34} \times 1.5 \times 10^{15} = 9.94 \times 10^{-19} \text{ J} KEmax⁡=E−W=4.64×10−19 JKE_{\max} = E - W = 4.64 \times 10^{-19} \text{ J} Vs=KEmax⁡e=4.64×10−191.6×10−19=2.9 VV_s = \frac{KE_{\max}}{e} = \frac{4.64 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.9 \text{ V}

Final Answer: 2.9 V

Example 9: First Line of Paschen Series

Calculate the wavelength of the first line in the Paschen series of hydrogen.

Solution: For Paschen series, nf=3n_f = 3. First line means ni=4n_i = 4. 1λ=RH(132−142)=1.097×107×7144\frac{1}{\lambda} = R_H\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = 1.097 \times 10^7 \times \frac{7}{144} 1λ=5.33×105 m−1\frac{1}{\lambda} = 5.33 \times 10^5 \text{ m}^{-1} λ=1.876×10−6 m=1876 nm\lambda = 1.876 \times 10^{-6} \text{ m} = 1876 \text{ nm}

Final Answer: 1876 nm

Example 10: Number of Spectral Lines

How many spectral lines are emitted when an electron transitions from n=5n=5 to n=1n=1 in hydrogen?

Solution: N=n(n−1)2=5×42=10N = \frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10

Final Answer: 10

Example 11: Energy and Wavelength of H-alpha Transition

Calculate the energy difference and wavelength for the transition n=3→n=2n=3 \to n=2 in hydrogen.

Solution: E3=−13.69=−1.51 eV,E2=−13.64=−3.4 eVE_3 = -\frac{13.6}{9} = -1.51 \text{ eV}, \quad E_2 = -\frac{13.6}{4} = -3.4 \text{ eV} ΔE=E3−E2=1.89 eV\Delta E = E_3 - E_2 = 1.89 \text{ eV} λ=12401.89=656 nm\lambda = \frac{1240}{1.89} = 656 \text{ nm}

Final Answer: 1.89 eV; 656 nm

Example 12: Lyman Series Limit

Calculate the wavelength of the series limit for the Lyman series of hydrogen.

Solution: At series limit, ni→∞n_i \to \infty, nf=1n_f = 1. 1λlimit=RH\frac{1}{\lambda_{\text{limit}}} = R_H λlimit=11.097×107=9.12×10−8 m=91.2 nm\lambda_{\text{limit}} = \frac{1}{1.097 \times 10^7} = 9.12 \times 10^{-8} \text{ m} = 91.2 \text{ nm} Energy: E=124091.2=13.6 eVE = \frac{1240}{91.2} = 13.6 \text{ eV}

Final Answer: 91.2 nm; 13.6 eV

Example 13: Radius of First Bohr Orbit

Calculate the radius of the first Bohr orbit in hydrogen.

Solution: rn=0.529×n2 A˚r_n = 0.529 \times n^2 \text{ Å} For n=1n=1: r1=0.529 A˚=5.29×10−11 mr_1 = 0.529 \text{ Å} = 5.29 \times 10^{-11} \text{ m}

Final Answer: 0.529 Å = 5.29×10−115.29 \times 10^{-11} m

Example 14: Velocity in Second Bohr Orbit

Calculate the electron velocity in the second Bohr orbit of hydrogen.

Solution: vn=2.19×106n m/sv_n = \frac{2.19 \times 10^6}{n} \text{ m/s} For n=2n=2: v2=1.095×106 m/sv_2 = 1.095 \times 10^6 \text{ m/s}

Final Answer: 1.095×1061.095 \times 10^6 m/s

Example 15: Ground-State Energy of He+^+

Calculate the energy of the first orbit of He+^+.

Solution: En=−13.6Z2n2 eVE_n = -\frac{13.6 Z^2}{n^2} \text{ eV} For He+^+, Z=2Z=2, n=1n=1: E1=−13.6×4=−54.4 eVE_1 = -13.6 \times 4 = -54.4 \text{ eV}

Final Answer: -54.4 eV

Example 16: Transition in Li2+^{2+}

Calculate the wavelength emitted when Li2+^{2+} transitions from n=3n=3 to n=2n=2.

Solution: 1λ=RHZ2(122−132)\frac{1}{\lambda} = R_H Z^2\left(\frac{1}{2^2} - \frac{1}{3^2}\right) For Z=3Z=3: 1λ=1.097×107×9×536=1.372×107 m−1\frac{1}{\lambda} = 1.097 \times 10^7 \times 9 \times \frac{5}{36} = 1.372 \times 10^7 \text{ m}^{-1} λ=7.29×10−8 m=72.9 nm\lambda = 7.29 \times 10^{-8} \text{ m} = 72.9 \text{ nm}

Final Answer: 72.9 nm

Example 17: de Broglie Wavelength of Electron

Calculate the de Broglie wavelength of an electron moving with velocity 2×1062 \times 10^6 m/s.

Solution: λ=hmv=6.626×10−349.109×10−31×2×106=3.64×10−10 m\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2 \times 10^6} = 3.64 \times 10^{-10} \text{ m}

Final Answer: 3.64×10−103.64 \times 10^{-10} m = 3.64 Å

Example 18: de Broglie Wavelength from Kinetic Energy

An electron has kinetic energy 100 eV. Calculate its de Broglie wavelength.

Solution: Using shortcut: λ=12.27KE(eV) A˚=12.2710=1.227 A˚\lambda = \frac{12.27}{\sqrt{KE(\text{eV})}} \text{ Å} = \frac{12.27}{10} = 1.227 \text{ Å}

Final Answer: 1.23 Å

Example 19: de Broglie Wavelength of Thermal Neutron

A thermal neutron has kinetic energy 0.025 eV. Calculate its de Broglie wavelength.

Solution: KE=0.025×1.6×10−19=4×10−21 JKE = 0.025 \times 1.6 \times 10^{-19} = 4 \times 10^{-21} \text{ J} λ=h2mKE=6.626×10−342×1.675×10−27×4×10−21\lambda = \frac{h}{\sqrt{2mKE}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 4 \times 10^{-21}}} λ=5.72×10−10 m\lambda = 5.72 \times 10^{-10} \text{ m}

Final Answer: 5.72×10−105.72 \times 10^{-10} m

Example 20: Electron vs Photon Wavelength at Same Energy

Compare the wavelengths of an electron and a photon, both having energy 10 eV.

Solution: For electron: λe=12.2710=3.88 A˚\lambda_e = \frac{12.27}{\sqrt{10}} = 3.88 \text{ Å} For photon: λγ=124010=124 nm\lambda_\gamma = \frac{1240}{10} = 124 \text{ nm} Ratio: λγλe=124×10−93.88×10−10≈31.9\frac{\lambda_\gamma}{\lambda_e} = \frac{124 \times 10^{-9}}{3.88 \times 10^{-10}} \approx 31.9

Final Answer: Electron wavelength = 3.88 Å; Photon wavelength = 124 nm; Photon wavelength is about 32 times larger

Example 21: Heisenberg Uncertainty Principle

An electron is confined to a region of space of width 1 Å. Calculate the minimum uncertainty in its velocity.

Solution: ΔxΔp≥h4π\Delta x \Delta p \ge \frac{h}{4\pi} Given Δx=10−10\Delta x = 10^{-10} m, Δp=6.626×10−344π×10−10=5.27×10−25 kg m/s\Delta p = \frac{6.626 \times 10^{-34}}{4\pi \times 10^{-10}} = 5.27 \times 10^{-25} \text{ kg m/s} Δv=Δpme=5.27×10−259.109×10−31=5.78×105 m/s\Delta v = \frac{\Delta p}{m_e} = \frac{5.27 \times 10^{-25}}{9.109 \times 10^{-31}} = 5.78 \times 10^5 \text{ m/s}

Final Answer: 5.78×1055.78 \times 10^5 m/s

Example 22: Neutron in Nucleus - Minimum Kinetic Energy

A neutron is confined within a nucleus of radius 10−1510^{-15} m. What is the minimum kinetic energy of the neutron?

Solution: Δp∼hΔx=6.626×10−3410−15=6.626×10−19 kg m/s\Delta p \sim \frac{h}{\Delta x} = \frac{6.626 \times 10^{-34}}{10^{-15}} = 6.626 \times 10^{-19} \text{ kg m/s} KEmin⁡=(Δp)22m=(6.626×10−19)22×1.675×10−27=1.31×10−10 JKE_{\min} = \frac{(\Delta p)^2}{2m} = \frac{(6.626 \times 10^{-19})^2}{2 \times 1.675 \times 10^{-27}} = 1.31 \times 10^{-10} \text{ J} In MeV: 1.31×10−101.6×10−13≈0.82 MeV\frac{1.31 \times 10^{-10}}{1.6 \times 10^{-13}} \approx 0.82 \text{ MeV}

Final Answer: 0.82 MeV

Example 23: Valid Quantum Number Sets

Identify which of the following sets (n,l,ml,ms)(n, l, m_l, m_s) are valid:

  1. (3, 2, -2, +1/2)
  2. (2, 2, 0, -1/2)
  3. (4, 3, +3, +1/2)
  4. (3, 2, +3, +1/2)

Solution: Rules:

  • 0≤l<n0 \le l < n
  • −l≤ml≤l-l \le m_l \le l
  • ms=±1/2m_s = \pm 1/2
  1. Valid
  2. Invalid (ll cannot equal nn)
  3. Valid
  4. Invalid (mlm_l cannot exceed ll)

Final Answer: 1 and 3 valid; 2 and 4 invalid

Example 24: Orbital Identification from Quantum Numbers

Name the orbitals having the following quantum numbers:

  1. (2, 0, 0)
  2. (3, 1, -1)
  3. (4, 3, 0)
  4. (3, 2, +2)

Solution: Using l=0→sl=0 \to s, l=1→pl=1 \to p, l=2→dl=2 \to d, l=3→fl=3 \to f:

  1. (2, 0, 0) → 2s
  2. (3, 1, -1) → 3p
  3. (4, 3, 0) → 4f
  4. (3, 2, +2) → 3d

Final Answer: 2s, 3p, 4f, 3d

Example 25: Energy of Photon in H-atom Transition

Calculate the energy of the photon emitted during a transition from n=5n=5 to n=2n=2 in hydrogen.

Solution: ΔE=2.18×10−18(122−152)=2.18×10−18×21100=4.58×10−19 J\Delta E = 2.18 \times 10^{-18}\left(\frac{1}{2^2} - \frac{1}{5^2}\right) = 2.18 \times 10^{-18} \times \frac{21}{100} = 4.58 \times 10^{-19} \text{ J}

Final Answer: 4.58×10−194.58 \times 10^{-19} J

Example 26: Angular Nodes in dx2−y2d_{x^2-y^2} Orbital

Find the number of angular nodes in dx2−y2d_{x^2-y^2} orbital.

Solution: Angular nodes = ll. For d-orbitals, l=2l=2.

Final Answer: 2

Example 27: Stability of Fe2+Fe^{2+} vs Fe3+Fe^{3+}

Which is more stable: Fe2+Fe^{2+} or Fe3+Fe^{3+}?

Solution: Fe2+=[Ar]3d6,Fe3+=[Ar]3d5Fe^{2+} = [Ar]3d^6, \quad Fe^{3+} = [Ar]3d^5 Fe3+Fe^{3+} has exactly half-filled d5d^5 configuration, which is extra stable due to symmetry and exchange energy.

Final Answer: Fe3+Fe^{3+}

Example 28: Magnetic Moment of Mn2+Mn^{2+}

Calculate the magnetic moment of Mn2+Mn^{2+}.

Solution: Mn2+=[Ar]3d5Mn^{2+} = [Ar]3d^5 Number of unpaired electrons n=5n=5. μ=n(n+2)=5(7)=35≈5.92 BM\mu = \sqrt{n(n+2)} = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \text{ BM}

Final Answer: 5.92 BM

Example 29: de Broglie Wavelength at c/10c/10

Find the wavelength of an electron moving with one-tenth the speed of light.

Solution: v=3×107 m/sv = 3 \times 10^7 \text{ m/s} λ=hmv=6.626×10−349.1×10−31×3×107=2.42×10−11 m\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 3 \times 10^7} = 2.42 \times 10^{-11} \text{ m}

Final Answer: 24.2 pm

Example 30: Frequency from Photon Energy

What is the frequency of a photon whose energy is 1 eV?

Solution: 1 eV=1.6×10−19 J1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} ν=Eh=1.6×10−196.626×10−34=2.41×1014 Hz\nu = \frac{E}{h} = \frac{1.6 \times 10^{-19}}{6.626 \times 10^{-34}} = 2.41 \times 10^{14} \text{ Hz}

Final Answer: 2.41×10142.41 \times 10^{14} Hz

Example 31: Number of Photons Emitted

Calculate the number of photons emitted by a 60 W source in 10 hours if the wavelength is 662.6 nm.

Solution: Total energy emitted: Etotal=60×10×3600=2.16×106 JE_{\text{total}} = 60 \times 10 \times 3600 = 2.16 \times 10^6 \text{ J} Energy of one photon: E=hcλ=6.626×10−34×3×108662.6×10−9≈3.00×10−19 JE = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{662.6 \times 10^{-9}} \approx 3.00 \times 10^{-19} \text{ J} Number of photons: N=2.16×1063.00×10−19=7.2×1024N = \frac{2.16 \times 10^6}{3.00 \times 10^{-19}} = 7.2 \times 10^{24}

Final Answer: 7.2×10247.2 \times 10^{24} photons

Example 32: Bohr Orbit as Standing Wave

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around it.

Solution: From Bohr's postulate: mvr=nh2πmvr = \frac{nh}{2\pi} Rearranging: 2πr=n(hmv)2\pi r = n\left(\frac{h}{mv}\right) Since λ=hmv\lambda = \frac{h}{mv}, 2πr=nλ2\pi r = n\lambda Thus, the orbit corresponds to a standing wave.

Final Answer: 2πr=nλ2\pi r = n\lambda