Question

Calculate the vibrational energy difference, in cm-1, between the v=0 and v=1 energy level in a hydrogen chloride-35 molecule. Isotope masses to three decimal places are taken from NIST and force constants from hyperphysics. Use the isotope mass, not the average mass.

Answer 1

For H-Cl³⁵ molecule we have,

Mass of H (m_H) = 1.007 amu, Mass of Cl-35 (m_Cl) = 34.968 amu.

Let us calculate reduced mass() of H-Cl moleule,

= (m_H) x (m_Cl) / (m_H) + (m_Cl)

= (1.007x34.968) / (1.007 + 34.968)

= 0.979 amu.

= 0.979 amu x (1.661 x 10-27) Kg/amu = 1.626 x 10^-27 Kg.

= 1.626 x 10^-27 Kg.

Bond force constant for H-Cl³⁵ is k = 481 N/m.

Now the Vibrational energy difference in cm^-1 units is calculated as follows

Formula, =(1/2c)x(k/)^1/2.

We have c = 2.998 x108 m/s, k = 491 N/m, = 1.626 x 10^-27 Kg.

Let us put all these values in eq.1 and solve it for   ,

= [1/(2x3.14x 2.998x10⁸)] / [491/(1.626 x 10^-27)]^1/2.

=5.311 x 10^-10 x 5.495 x 10¹⁴.

= 2.918 x 10⁵ m^-1.

= 2.918 x 10⁵ x 10^-2 cm^-1.....(1 m = 10² cm i.e. 1 m^-1 = 10^-2 cm-1

= 2.918 x 10³ cm^-1.

Hence for H³⁵Cl molecule Vibrational energy level difference is 2.918 x 10³ cm^-1.