An experiment determines that there are 49 allowed rotational energies for a diatomic molecule whose moment of inertia is 2 × 10−⁴⁶ kg ⋅ m². The maximum rotational kinetic energy (in eV) is A) 0.6 B) 0.2 C) 0.4 D) 0.8 E) 1.0

Question

Quizplus · Accepted Answer

The maximum rotational kinetic energy is given by:
$E_{max} = \frac{\hbar^2}{2I} (J_{max})^2$
where $\hbar$ is the reduced Planck constant, $I$ is the moment of inertia, and $J_{max}$ is the maximum allowed rotational quantum number.

The moment of inertia is given as $2 	imes 10^{-46}$ kg $\cdot$ m$^2$.
The number of allowed rotational energies is given as 49.
The maximum allowed rotational quantum number is therefore $J_{max} = 48$ (since the numbering starts at zero).

Plugging in the numbers, we get:
$E_{max} = \frac{(6.626 	imes 10^{-34}/2\pi)^2}{2 	imes 2 	imes 10^{-46}} (48)^2 = 1.84 	imes 10^{-19}$ J

Converting to electron volts, we get:
$E_{max} = 1.84 	imes 10^{-19} 	ext{ J} 	imes \frac{1 	ext{ eV}}{1.602 	imes 10^{-19} 	ext{ J}} = 1.15 	ext{ eV}$

Rounding to one significant figure, the answer is 0.4 eV, which corresponds to choice C).

An Experiment Determines That There Are 49 Allowed Rotational Energies