Question

A lithium atom has mass 1.17×10−26kg, and a hydrogen atom has mass 1.67×10−27kg. The equilibrium separation between the two nuclei in the LiH molecule is 0.159 nm.A. What is the difference in energy between the l = 3 and l = 4 rotational levels?B. What is the wavelength of the photon emitted in a transition from the l = 4 to l = 3 level?

Answer 1

The difference in energy between rotational levels can be found using the formula ΔE = (h^2/2I)(l(l+1)-l'(l'+1)). The wavelength of the photon emitted in a transition between levels can be calculated using ΔE = hc/λ.

Step-by-step explanation:

To find the difference in energy between the l=3 and l=4 rotational levels, we can use the formula ΔE = (h^2/2I)(l(l+1)-l'(l'+1)), where ΔE is the difference in energy, h is Planck's constant (6.626 x 10^-34 J s), I is the moment of inertia, and l and l' are the initial and final rotational quantum numbers, respectively.

For a diatomic molecule like LiH, the moment of inertia can be approximated by I = µR^2, where µ is the reduced mass of the system and R is the equilibrium separation between the two nuclei.

The wavelength of the photon emitted in a transition from the l=4 to l=3 level can be calculated using the formula ΔE = hc/λ, where ΔE is the difference in energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

Answer 2

Step-by-step explanation:

To calculate the energy difference between rotational levels and the wavelength of the emitted photon in a transition, we can use the rotational energy formula for a diatomic molecule:

E = (l(l + 1)ħ²)/(2I)

where E is the energy, l is the rotational quantum number, ħ is the reduced Planck's constant (h/2π), and I is the moment of inertia of the molecule.

Given:

Mass of lithium atom (m₁) = 1.17 × 10^(-26) kg

Mass of hydrogen atom (m₂) = 1.67 × 10^(-27) kg

Equilibrium separation (r) = 0.159 nm = 0.159 × 10^(-9) m

(a) Energy difference between l = 3 and l = 4 rotational levels:

First, we need to calculate the moment of inertia (I) of the LiH molecule. The moment of inertia can be approximated as the reduced mass (μ) multiplied by the equilibrium separation (r) squared:

μ = (m₁ * m₂) / (m₁ + m₂) [Reduced mass]

I = μ * r² [Moment of inertia]

Calculating the reduced mass (μ):

μ = (1.17 × 10^(-26) kg * 1.67 × 10^(-27) kg) / (1.17 × 10^(-26) kg + 1.67 × 10^(-27) kg)

= 1.95 × 10^(-28) kg

Calculating the moment of inertia (I):

I = (1.95 × 10^(-28) kg) * (0.159 × 10^(-9) m)²

= 4.93 × 10^(-47) kg·m²

Now we can calculate the energy for l = 3 and l = 4:

E₃ = (3(3 + 1)ħ²)/(2I)

E₄ = (4(4 + 1)ħ²)/(2I)

The energy difference is:

ΔE = E₄ - E₃

Substitute the value of the reduced Planck's constant, ħ = 6.626 × 10^(-34) J·s, and perform the calculations to find the energy difference.

(b) Wavelength of the photon emitted in a transition from l = 4 to l = 3:

The energy difference between two rotational levels can be used to calculate the wavelength of the emitted photon using the equation:

ΔE = hc/λ

where ΔE is the energy difference, h is Planck's constant (6.626 × 10^(-34) J·s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength of the photon.

We can rearrange the equation to solve for the wavelength:

λ = hc/ΔE

Substitute the value of ΔE (energy difference) calculated in part (a), and perform the calculations to find the wavelength of the emitted photon.