The internuclear distance (bond length) of carbon monoxide molecule is 1.13 Ã…. Calculate the energy (in joules and eV) of this molecule in the first excited rotational level. Also calculate the angular velocity of the molecule. Given atomic masses of 12^C = 1.99×10^-26 kg; 16^O = 2.66×10^-26 kg.
The Correct Answer and Explanation is :
To calculate the energy of the carbon monoxide (CO) molecule in its first excited rotational level, we need to use the rotational energy formula for a rigid rotor:
[
E_J = \frac{J(J + 1)h^2}{8\pi^2I}
]
Where:
- ( E_J ) is the rotational energy in the (J^{th}) level.
- ( J ) is the rotational quantum number (for the first excited state, ( J = 1 )).
- ( h ) is Planck’s constant (( h = 6.626 \times 10^{-34} \, \text{J·s} )).
- ( I ) is the moment of inertia of the molecule, defined as:
[
I = \mu r^2
]
Where:
- ( \mu ) is the reduced mass of the molecule,
- ( r ) is the bond length (1.13 Å = ( 1.13 \times 10^{-10} \, \text{m} )).
The reduced mass of the CO molecule is calculated using:
[
\mu = \frac{m_C m_O}{m_C + m_O}
]
Given:
- The mass of ( ^{12}C ) (carbon) is ( 1.99 \times 10^{-26} \, \text{kg} ),
- The mass of ( ^{16}O ) (oxygen) is ( 2.66 \times 10^{-26} \, \text{kg} ).
Step 1: Calculate the reduced mass (( \mu ))
[
\mu = \frac{(1.99 \times 10^{-26} \, \text{kg})(2.66 \times 10^{-26} \, \text{kg})}{1.99 \times 10^{-26} + 2.66 \times 10^{-26}}
]
[
\mu = 1.22 \times 10^{-26} \, \text{kg}
]
Step 2: Calculate the moment of inertia (( I ))
[
I = \mu r^2 = (1.22 \times 10^{-26} \, \text{kg})(1.13 \times 10^{-10} \, \text{m})^2
]
[
I = 1.57 \times 10^{-46} \, \text{kg·m}^2
]
Step 3: Calculate the rotational energy for ( J = 1 )
[
E_1 = \frac{(1)(1 + 1)(6.626 \times 10^{-34})^2}{8\pi^2 (1.57 \times 10^{-46})}
]
[
E_1 = 1.39 \times 10^{-23} \, \text{J}
]
Step 4: Convert energy from joules to electron volts (eV)
To convert the energy from joules to eV:
[
1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}
]
[
E_1 = \frac{1.39 \times 10^{-23}}{1.602 \times 10^{-19}} = 8.69 \times 10^{-5} \, \text{eV}
]
Step 5: Calculate the angular velocity (( \omega ))
The angular velocity is given by:
[
\omega = \frac{E_1}{I}
]
[
\omega = \frac{1.39 \times 10^{-23}}{1.57 \times 10^{-46}} = 8.86 \times 10^{22} \, \text{rad/s}
]
Final Results:
- The energy of the CO molecule in its first excited rotational level is approximately ( 1.39 \times 10^{-23} \, \text{J} ) or ( 8.69 \times 10^{-5} \, \text{eV} ).
- The angular velocity of the molecule is ( 8.86 \times 10^{22} \, \text{rad/s} ).
Explanation:
The rotational energy levels of a diatomic molecule are quantized and depend on the moment of inertia, which is related to the reduced mass and bond length of the molecule. In this case, we used the values of the masses of carbon and oxygen atoms, the bond length of CO, and Planck’s constant to determine the rotational energy in the first excited state. The reduced mass was calculated by combining the masses of the two atoms involved. The energy was then converted to eV to express it in a more familiar unit. The angular velocity was found from the rotational energy and moment of inertia, giving an idea of how fast the molecule rotates at this energy level.