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

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 (h)/(2), i.e., mvr = n(h)/(2). • 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 E = E_2 - E_1 = h. ii) Shortcomings of Bohr's atomic model: • It could only explain the spectra of hydrogen and hydrogen-like ions (e.g., He^+, 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. b) Step 1: Identify the series and transition for the longest wavelength. For the Balmer series, the final energy level is n_f = 2. For the longest wavelength (smallest energy difference), the electron transitions from the next highest energy level, which is n_i = 3. Step 2: Use the Rydberg formula to calculate the wave number. The Rydberg formula for the wave number () is: = R_H ( (1)/(n_f^2) - (1)/(n_i^2) ) where R_H = 1.097 × 10^7 m^-1 is the Rydberg constant. Step 3: Substitute the values and calculate. = (1.097 × 10^7 m^-1) ( (1)/(2^2) - (1)/(3^2) ) = (1.097 × 10^7 m^-1) ( (1)/(4) - (1)/(9) ) = (1.097 × 10^7 m^-1) ( (9-4)/(36) ) = (1.097 × 10^7 m^-1) ( (5)/(36) ) = 1.5236 × 10^6 m^-1 The wave number of the longest wavelength transition in the Balmer series is 1.524 × 10^6 m^-1 c) i) Step 1: Identify the given values. Planck's constant, h = 6.626 × 10^-34 J s Mass of neutron, m = 1.614 × 10^-27 Kg Speed of light, c = 3 × 10^8 m/s Velocity of neutron, v = 0.05 × c = 0.05 × (3 × 10^8 m/s) = 1.5 × 10^7 m/s Step 2: Apply the de Broglie wavelength formula. The de Broglie wavelength () is given by: = (h)/(mv) Step 3: Substitute the values and calculate. = 6.626 × 10^-34 J s(1.614 × 10^-27 Kg) × (1.5 × 10^7 m/s) = 6.626 × 10^-342.421 × 10^-20 m = 2.737 × 10^-14 m The de Broglie wavelength for the neutron is 2.737 × 10^-14 m ii) Water (H_2O) is liquid at room temperature while hydrogen sulfide (H_2S) is a gas, despite H_2S having a higher molecular weight. This is due to the presence of strong hydrogen bonding in H_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 H_2S does not form hydrogen bonds. Instead, H_2S molecules are held together by weaker dipole-dipole interactions and London dispersion forces. The stronger hydrogen bonds in H_2O require more energy to overcome, resulting in a higher boiling point and thus H_2O being liquid at room temperature.