Atomic Structure MCQs Question and Answer in Engineering Chemistry

This question is about the de Broglie wavelength, which tells us that even particles like atoms can behave like waves!
The de Broglie wavelength (λ) is calculated using the formula: λ = h/mv
where:
h is Planck's constant (6.626 x 10^-34 Js)
m is the mass of the particle (in kg)
v is the velocity of the particle (in m/s)
First, we need the mass of a Helium (He) atom. A He atom has a mass of approximately 4 amu (atomic mass units). To use the formula, we need to convert this to kilograms. Since 1 amu ≈ 1.66 x 10^-27 kg, the mass of a He atom is roughly 4 * 1.66 x 10^-27 kg = 6.64 x 10^-27 kg.
Now, we can plug the values into the de Broglie wavelength formula:
λ = (6.626 x 10^-34 Js) / (6.64 x 10^-27 kg * 1000 m/s)
After calculating, you will get a wavelength around 99.7 x 10^-12 m.

The vibrational partition function describes how molecules are distributed among different vibrational energy levels at thermal equilibrium.

For a molecule that behaves as a quantum mechanical simple harmonic oscillator, the vibrational energy levels are given by E_v = (v + 1/2)hν, where v = 0, 1, 2, ....

When the zero-point energy is treated separately, the vibrational partition function becomes:

q_vib = 1 / [1 − exp(−hν/k_BT)]

This expression is equivalent to:

q_vib = [1 − exp(−hν/k_BT)]⁻¹

Therefore, Option B is the correct expression for the vibrational partition function.

Why the other options are incorrect:

Option A: It contains only the exponential term and is not the complete partition function.

Option C: It includes an unnecessary extra exponential factor, so it does not represent the standard vibrational partition function.

Option D: It contains the factor exp(−hν/2k_BT), which accounts for the zero-point energy. This form is obtained only when the zero-point energy is explicitly included in the partition function and is not the standard expression generally used in engineering chemistry and statistical thermodynamics.