Solution Analysis

This question asks for the wavenumber of the radiation required to cause a rotational transition from the J = 0 state to the J = 1 state in a diatomic molecule (XY). This calculation relies on the rigid rotor model and requires the molecule's moment of inertia, derived from its reduced mass and bond length.

Key Concepts and Formulas

• The rotational energy levels of a diatomic molecule, treated as a rigid rotor, are given by:
  $E_J = \frac{h^2}{8\pi^2 I} J(J+1)$ where E_J is the energy of the rotational level, h is Planck's constant, I is the moment of inertia, and J is the rotational quantum number (J = 0, 1, 2, ...).
• The selection rule for rotational transitions requires that the change in the rotational quantum number is $\Delta J = \pm 1$. For absorption of radiation, $\Delta J = +1$.
• The transition considered is from $J=0$ to $J=1$.
• The energy difference ($\Delta E$) for this transition is:
  $\Delta E = E_1 - E_0 = \left( \frac{h^2}{8\pi^2 I} (1)(1+1) \right) - \left( \frac{h^2}{8\pi^2 I} (0)(0+1) \right) = \frac{h^2}{4\pi^2 I}$.
• The energy absorbed is equal to the energy of the photon: $\Delta E = h\nu$, where $\nu$ is the frequency of the radiation.
• Equating the two expressions for $\Delta E$:
  $h\nu = \frac{h^2}{4\pi^2 I}$
  $\nu = \frac{h}{4\pi^2 I}$.
• The wavenumber ($\tilde{\nu}$) is related to frequency by $\tilde{\nu} = \frac{\nu}{c}$, where $c$ is the speed of light. Therefore, the formula for wavenumber is:
  $\tilde{\nu} = \frac{h}{4\pi^2 I c}$. This formula gives $\tilde{\nu}$ in units of m$^{-1}$.
• The moment of inertia (I) for a diatomic molecule is calculated using the reduced mass ($\mu$) and the bond length ($r$):
  $I = \mu r^2$.

The force constant provided relates to vibrational frequency and is not used in this calculation for rotational excitation.

Calculation Steps

1. Calculate the Moment of Inertia (I):
   • Given reduced mass, $\mu = 2 \times 10^{-26}$ kg.
   • Given bond length, $r = 100$ pm $= 100 \times 10^{-12}$ m $= 1 \times 10^{-10}$ m.
   • $I = \mu r^2 = (2 \times 10^{-26} \text{ kg}) \times (1 \times 10^{-10} \text{ m})^2$
   • $I = 2 \times 10^{-26} \times 1 \times 10^{-20} \text{ kg m}^2 = 2 \times 10^{-46} \text{ kg m}^2$.
2. Calculate the Wavenumber ($\tilde{\nu}$) in m$^{-1}$:
   • Use the physical constants: Planck's constant, $h \approx 6.626 \times 10^{-34}$ J s; speed of light, $c \approx 3 \times 10^8$ m s$^{-1}$. Use $\pi \approx 3.14159$.
   • Substitute the values into the formula:
     $\tilde{\nu} = \frac{6.626 \times 10^{-34} \text{ J s}}{4\pi^2 \times (2 \times 10^{-46} \text{ kg m}^2) \times (3 \times 10^8 \text{ m s}^{-1})}$
   • Calculate the denominator: $4\pi^2 \times (2 \times 10^{-46}) \times (3 \times 10^8) \approx (39.478) \times (6 \times 10^{-38}) \approx 236.87 \times 10^{-38} \approx 2.3687 \times 10^{-36} \text{ kg m}^3 \text{ s}^{-1}$.
   • $\tilde{\nu} \approx \frac{6.626 \times 10^{-34}}{2.3687 \times 10^{-36}} \text{ m}^{-1}$
   • $\tilde{\nu} \approx 2.797 \times 10^2 \text{ m}^{-1} \approx 280 \text{ m}^{-1}$.
3. Convert Wavenumber to cm$^{-1}$:
   • Divide the result in m$^{-1}$ by 100 (since 1 m = 100 cm).
   • $\tilde{\nu} (\text{cm}^{-1}) = \frac{280 \text{ m}^{-1}}{100 \text{ cm/m}} = 2.8 \text{ cm}^{-1}$.

Conclusion

The wavenumber of radiation required for the $J=0 \to J=1$ rotational transition is calculated to be 2.8 cm$^{-1}$.