Question

The fundamental vibrational frequency of \( ^1H^{127}I \) is 2309 cm\(^{-1}\). The force constant for this molecule (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\) N m\(^{-1}\).
Given: \( N = 6.022 \times 10^{23}, c = 3.0 \times 10^8 \, \text{m s}^{-1} \).

Hint:

The force constant for a molecule can be calculated using the vibrational frequency and the reduced mass of the atoms involved in the vibration.

Answer 1

Correct Answer: 309

The force constant \( k \) can be related to the vibrational frequency \( \nu \) by the formula:
\[ k = \mu \nu^2 \] where \( \mu \) is the reduced mass of the system and \( \nu \) is the frequency. For \( ^1H^{127}I \), we can calculate the reduced mass \( \mu \) as:
\[ \mu = \frac{m_1 m_2}{m_1 + m_2}. \] The atomic masses of hydrogen and iodine are 1 and 127 amu, respectively.
\[ \mu = \frac{1 \times 127}{1 + 127} = 0.992 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg}. \] Now, we can substitute the values into the formula for \( k \), where \( \nu = 2309 \, \text{cm}^{-1} \) and converting to m\(^{-1}\):
\[ \nu = 2309 \times 100 = 2.309 \times 10^5 \, \text{m}^{-1}. \] Thus, the force constant is:
\[ k = (1.66 \times 10^{-27}) \times (2.309 \times 10^5)^2 \approx 18245 \, \text{N m}^{-1}. \] Thus, the force constant is \( 18245 \, \text{N m}^{-1} \).