Question

The ratio of the fundamental vibrational frequencies \( \left( \nu_{^{13}C^{16}O} / \nu_{^{12}C^{16}O} \right) \) of two diatomic molecules \( ^{13}C^{16}O \) and \( ^{12}C^{16}O \), considering their force constants to be the same, is \underline{{2cm} (rounded off to two decimal places).}

Hint:

For diatomic molecules, the vibrational frequency is inversely proportional to the square root of the reduced mass. Heavier isotopes result in a lower frequency.

Answer 1

Correct Answer: 0.97

For diatomic molecules, the vibrational frequency is given by: \[ \nu \propto \frac{1}{\sqrt{\mu}}, \quad {where } \mu = \frac{m_1 m_2}{m_1 + m_2} \] Calculate reduced masses (in atomic mass units): \[ \mu_{^{12}C^{16}O} = \frac{12 \times 16}{12 + 16} = \frac{192}{28} = 6.8571 \] \[ \mu_{^{13}C^{16}O} = \frac{13 \times 16}{13 + 16} = \frac{208}{29} = 7.1724 \] Now compute the ratio of frequencies: \[ \frac{\nu_{^{13}C^{16}O}}{\nu_{^{12}C^{16}O}} = \sqrt{\frac{\mu_{^{12}C^{16}O}}{\mu_{^{13}C^{16}O}}} = \sqrt{\frac{6.8571}{7.1724}} \approx \sqrt{0.9560} \approx 0.9778 \] \[ \boxed{0.98} \]