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Explain Bohr's atomic model postulates and shortcomings. Calculate the longest wavelength in the Balmer series.

Physics

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2 parts
b

1.524×106 m−1\text{}1.524 \times 10^6 \text{ m}^{-1}\text{}

c(i)

2.737×10−14 m\text{}2.737 \times 10^{-14} \text{ m}\text{}

Step-by-step solution

6 steps

Part a(i)

Postulates of Bohr's atomic model

  • Electrons revolve around the nucleus in specific circular orbits called stationary states without radiating energy.
  • Electrons can only exist in orbits where their angular momentum is an integral multiple of h2π\frac{h}{2\pi}, i.e., mvr=nh2πmvr = n\frac{h}{2\pi}.
  • Energy is absorbed when an electron jumps from a lower energy orbit to a higher energy orbit, and energy is emitted when it jumps from a higher energy orbit to a lower energy orbit. The energy difference is given by ΔE=E2−E1=hν\Delta E = E_2 - E_1 = h\nu.

Part a(ii)

Shortcomings of Bohr's atomic model

  • It could only explain the spectra of hydrogen and hydrogen-like ions (e.g., He+He^+, Li2+Li^{2+}) but failed for multi-electron atoms.
  • It could not explain the splitting of spectral lines into finer lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
  • It did not account for the wave nature of electrons (de Broglie hypothesis) or the Heisenberg uncertainty principle.

Part b

1.524×106 m−1\text{}1.524 \times 10^6 \text{ m}^{-1}\text{}

  1. Identify the series and transition for the longest wavelength.

    For the Balmer series, the final energy level is nf=2n_f = 2.
    For the longest wavelength (smallest energy difference), the electron transitions from the next highest energy level, which is ni=3n_i = 3.

  2. Use the Rydberg formula to calculate the wave number.

    The Rydberg formula for the wave number (νˉ\bar{\nu}) is:

    νˉ=RH(1nf2−1ni2)\bar{\nu} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

    where RH=1.097×107 m−1R_H = 1.097 \times 10^7 \text{ m}^{-1} is the Rydberg constant.

  3. Substitute the values and calculate.

    νˉ=(1.097×107 m−1)(122−132)\bar{\nu} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) νˉ=(1.097×107 m−1)(14−19)\bar{\nu} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{1}{4} - \frac{1}{9} \right) νˉ=(1.097×107 m−1)(9−436)\bar{\nu} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{9-4}{36} \right) νˉ=(1.097×107 m−1)(536)\bar{\nu} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{5}{36} \right) νˉ=1.5236×106 m−1\bar{\nu} = 1.5236 \times 10^6 \text{ m}^{-1}

    The wave number of the longest wavelength transition in the Balmer series is 1.524×106 m−1\boxed{\text{}1.524 \times 10^6 \text{ m}^{-1}\text{}}

Part c(i)

2.737×10−14 m\text{}2.737 \times 10^{-14} \text{ m}\text{}

  1. Identify the given values.

    Planck's constant, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}
    Mass of neutron, m=1.614×10−27 Kgm = 1.614 \times 10^{-27} \text{ Kg}
    Speed of light, c=3×108 m/sc = 3 \times 10^8 \text{ m/s}
    Velocity of neutron, v=0.05×c=0.05×(3×108 m/s)=1.5×107 m/sv = 0.05 \times c = 0.05 \times (3 \times 10^8 \text{ m/s}) = 1.5 \times 10^7 \text{ m/s}

  2. Apply the de Broglie wavelength formula.

    The de Broglie wavelength (λ\lambda) is given by:

    λ=hmv\lambda = \frac{h}{mv}
  3. Substitute the values and calculate.

    λ=6.626×10−34 J s(1.614×10−27 Kg)×(1.5×107 m/s)\lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{(1.614 \times 10^{-27} \text{ Kg}) \times (1.5 \times 10^7 \text{ m/s})} λ=6.626×10−342.421×10−20 m\lambda = \frac{6.626 \times 10^{-34}}{2.421 \times 10^{-20}} \text{ m} λ=2.737×10−14 m\lambda = 2.737 \times 10^{-14} \text{ m}

    The de Broglie wavelength for the neutron is 2.737×10−14 m\boxed{\text{}2.737 \times 10^{-14} \text{ m}\text{}}

Part c(ii)

Water (H2OH_2O) is liquid at room temperature while hydrogen sulfide (H2SH_2S) is a gas, despite H2SH_2S having a higher molecular weight. This is due to the presence of strong hydrogen bonding in H2OH_2O. Oxygen is highly electronegative, leading to highly polar O-H bonds and allowing hydrogen atoms to form strong intermolecular hydrogen bonds. Sulfur is less electronegative than oxygen, so H2SH_2S does not form hydrogen bonds. Instead, H2SH_2S molecules are held together by weaker dipole-dipole interactions and London dispersion forces. The stronger hydrogen bonds in H2OH_2O require more energy to overcome, resulting in a higher boiling point and thus H2OH_2O being liquid at room temperature.

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