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The vibrational energy of a harmonic oscillator is given by Schrodinger's relation. $$ { E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) h\omega \qquad \qquad \qquad ...(1)$$ where \(v\) = vibrational quantum number having values 0, 1, 2, ..... etc.
\(h\) = Planck's constant
\( \omega\) = fundamental frequency of vibration $$ \omega = \frac { 1 }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$ Lets put this value in equation (1), we get, $$ { E }_{ v } = \left( v + \frac { 1 }{ 2 } \right) \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$ For lowest energy level, \(v = 0\) $$ { E }_{ v } = \left( 0 + \frac { 1 }{ 2 } \right) \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$ $$ { E }_{ v } = \frac { 1 }{ 2 } \frac { h }{ 2\pi } \sqrt { \frac { K }{ \mu } } $$ This energy possessed by molecule when \(v = 0\) is called Zero Point Energy. The zero point energy indicates that the atoms can never be at rest, the molecules must vibrate even at \(0\) K. i.e., at absolute zero, where translational and rotational energies are to be zero.
The vibrational frequency of HCl is \(2.988 \times { 10 }^{ 5 } { m }^{ -1 }\). Calculate the zero point energy of the molecule.
Solution: We have wave number \(\overline { \upsilon } = 2.988 \times { 10 }^{ 5 } { m }^{ -1 }\)
Zero Point Energy is the vibrational energy at \(v = 0\) $$ \therefore { E }_{ 0 } = \frac { 1 }{ 2 } h\omega = \frac { 1 }{ 2 } h\upsilon = \frac { 1 }{ 2 } h\overline { \upsilon } c $$ $$ = \frac { 1 }{ 2 } \left( 6.626 \times { 10 }^{ -34 } Js \right) \left( 2.988 \times { 10 }^{ 5 } { m }^{ -1 } \right) \left( 3 \times { 10 }^{ 8 } m/s \right) $$ $$ \qquad = \frac { 1 }{ 2 } \left( 5.94 \times { 10 }^{ -20 } \right) $$ $$ \qquad = 2.97 \times { 10 }^{ -20 } J $$
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