For H2 molecule, the fundamental vibrational frequency (ve ) in wavenumbers can be taken as 4400 cm-1. The zero-point energy of the molecule is ____ kJ mol-1 (rounded up to two decimal places). [h=6.6 10-34 Js, c=3 108 ms-1, NA=6 1023 mol-1 ]

ABC26PH0766 · Physical Spectroscopy

Subject: Physical Chemistry · Chapter: Physical Spectroscopy · Topic: Physical Spectroscopy – General · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Medium

For $\mathrm{H}_{2}$ molecule, the fundamental vibrational frequency $\left(\bar{v}_{\mathrm{e}}\right)$ in wavenumbers can be taken as $4400 \mathrm{cm}^{-1}$. The zero-point energy of the molecule is ____ $\mathrm{kJ} \mathrm{mol}^{-1}$ (rounded up to two decimal places).

\[ \left[\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3 \times 10^{8} \mathrm{ms}^{-1}, \mathrm{N}_{\mathrm{A}}=6 \times 10^{23} \mathrm{mol}^{-1}\right] \]

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