Question

The force constant for H$^{35}$Cl and D$^{35}$Cl are the same and both can be considered as harmonic oscillators. H$^{35}$Cl has a fundamental vibrational transition at 2886 cm$^{-1}$. The ratio of the zero-point energy of H$^{35}$Cl to that of D$^{35}$Cl is
 

• A. 0.515
• B. 0.717
• C. 1.395
• D. 1.946

Hint:

For harmonic oscillators, the zero-point energy is inversely related to the square root of the reduced mass. As deuterium (D) is heavier than hydrogen (H), the zero-point energy for D$^{35}$Cl is lower than for H$^{35}$Cl.

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the ratio of the zero-point energy of H³⁵Cl to that of D³⁵Cl, given that both can be considered as harmonic oscillators and the force constants for both molecules are the same.

The fundamental frequency transition, \(\nu_0\), of the harmonic oscillator is related to the following formula:

\(\nu_0 = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}}\)

where:

• \(k\) is the force constant (same for both molecules).
• \(\mu\) is the reduced mass of the oscillator.
• \(c\) is the speed of light.

The zero-point energy (ZPE) of a harmonic oscillator is given by:

\(\text{ZPE} = \frac{1}{2} h \nu\)

For H³⁵Cl and D³⁵Cl:

\(\nu_{\text{HCl}} = 2886 \text{ cm}^{-1}\)

Since the force constant is the same, the ratio of frequencies is inversely proportional to the square root of their reduced masses:

\(\frac{\nu_{\text{HCl}}}{\nu_{\text{DCl}}} = \sqrt{\frac{\mu_{\text{DCl}}}{\mu_{\text{HCl}}}}\)

To find the reduced mass, we use the formula:

\(\mu = \frac{m_{\text{Cl}} \cdot m_{\text{X}}}{m_{\text{Cl}} + m_{\text{X}}}\)

where \(m_{\text{Cl}}\) is the mass of Cl and \(m_{\text{X}}\) is the mass of H or D.

Given atomic masses in atomic mass units (amu):

• \(m_{\text{H}} = 1 \text{ amu}\)
• \(m_{\text{D}} = 2 \text{ amu}\)
• \(m_{\text{Cl}} = 35 \text{ amu}\)

Reduced mass, \(\mu_{\text{HCl}}\):

\(\mu_{\text{HCl}} = \frac{35 \times 1}{35 + 1} = \frac{35}{36}\)

Reduced mass, \(\mu_{\text{DCl}}\):

\(\mu_{\text{DCl}} = \frac{35 \times 2}{35 + 2} = \frac{70}{37}\)

Now, calculate the ratio of zero-point energy:

\(\frac{\text{ZPE}_{\text{HCl}}}{\text{ZPE}_{\text{DCl}}} = \frac{\nu_{\text{HCl}}}{\nu_{\text{DCl}}} = \sqrt{\frac{\mu_{\text{DCl}}}{\mu_{\text{HCl}}}}\)

Substitute the values:

\(\frac{\text{ZPE}_{\text{HCl}}}{\text{ZPE}_{\text{DCl}}} = \sqrt{\frac{70}{37} \times \frac{36}{35}}\)

Calculate the final ratio:

\(\sqrt{\frac{70 \times 36}{37 \times 35}} \approx 1.395\)

Therefore, the correct answer is 1.395.