Question

What is the difference in energy ΔE between the n=0 and n=1 vibrational states for diatomic hydrogen? The effective force constant k for diatomic hydrogen is 576N/m .

Answer 1

The energy difference (
`\( \Delta E \)`) between the n=0 and n=1 vibrational states for diatomic hydrogen is approximately
`\(9.35 * 10^(-21) \, \text{J}\).`

The energy difference (
`\( \Delta E \)`) between vibrational states in a diatomic molecule can be calculated using the formula for harmonic oscillator energy levels:

`\[ \Delta E = h \cdot \\u \]`

where:

- h is Planck's constant (
`\(6.626 * 10^(-34) \, \text{J} \cdot \text{s}\)`),

-
`\( \\u \)` is the vibrational frequency.

For a diatomic molecule, the vibrational frequency (
`\( \\u \)`) can be calculated using the formula:

`\[ \\u = (1)/(2\pi) \sqrt{(k)/(\mu)} \]`

where:

- k is the force constant,

-
`\( \mu \)` is the reduced mass of the system.

The reduced mass (
`\( \mu \)`) for a diatomic molecule can be calculated using:

`\[ \mu = (m_1 \cdot m_2)/(m_1 + m_2) \]`

where:

-
`\( m_1 \) and \( m_2 \)` are the masses of the two atoms.

For diatomic hydrogen (
`\( H_2 \)`), each hydrogen atom has a mass of approximately
`\(1.0079 \, \text{u}\)` (atomic mass units).

Now, let's plug in the values:

`\[ \mu = \frac{(1.0079 \, \text{u})^2}{1.0079 \, \text{u} + 1.0079 \, \text{u}} \]`

`\[ \mu \approx 0.5039 \, \text{u} \]`

`\[ \\u = (1)/(2\pi) \sqrt{\frac{576 \, \text{N/m}}{0.5039 \, \text{u}}} \]`

`\[ \\u \approx 1.412 * 10^(13) \, \text{Hz} \]`

Now, calculate
`\( \Delta E \)`:

`\[ \Delta E = h \cdot \\u \]`

`\[ \Delta E \approx (6.626 * 10^(-34) \, \text{J} \cdot \text{s}) \cdot (1.412 * 10^(13) \, \text{Hz}) \]`

`\[ \Delta E \approx 9.35 * 10^(-21) \, \text{J} \]`

Therefore, the energy difference (
`\( \Delta E \)`) between the n=0 and n=1 vibrational states for diatomic hydrogen is approximately
`\(9.35 * 10^(-21) \, \text{J}\).`