A LiBr molecule oscillates with a frequency of 1.7×1013 Hz.

(a) What is the difference in energy in eV between allowed oscillator states?
(b) What is the approximate value of n for a state having an energy of 1.0 eV?

E=hf=6.63•10⁻³⁴•1.7•10¹³=1.13•10⁻²⁰ J=

= (1.13•10⁻²⁰/1.6•10⁻¹⁹) eV = 0.07 eV
E=(n+0.5)hf
n=(E/hf) - 0.5 =
=(1.6•10⁻¹⁹/6.63•10⁻³⁴•1.7•10¹³) – 0.5 ≃14

(a) Oh boy, we're talking about energy differences! I hope you're ready for a shockingly electrifying answer! The energy of an oscillator state is given by the formula E = (n + 1/2)hf, where n is the principal quantum number, h is Planck's constant (approximately 4.1357 x 10^-15 eV·s), and f is the frequency. So, let's plug in the values and calculate, shall we?

E₁ = (n₁ + 1/2)hf
E₂ = (n₂ + 1/2)hf

The difference in energy (ΔE) between two allowed oscillator states can be obtained by subtracting the energy of one state from the other:

ΔE = E₂ - E₁

Now, we know the frequency (f) is 1.7 x 10^13 Hz. So, let's calculate, clown-style:

ΔE = ((n₂ + 1/2)hf) - ((n₁ + 1/2)hf)
= (n₂ - n₁)hf

Substituting the values into the equation, we get:

ΔE = (n₂ - n₁)hf
= (n₂ - n₁) × 4.1357 x 10^-15 eV·s × 1.7 x 10^13 Hz

And there you have it, the difference in energy between allowed oscillator states for a LiBr molecule!

(b) Now, let's find out the approximate value of n for a state having an energy of 1.0 eV. Remember our formula E = (n + 1/2)hf? We can rearrange that to solve for n:

n = (E/hf) - 1/2

Substituting the given values:

n = (1.0 eV / (4.1357 x 10^-15 eV·s × 1.7 x 10^13 Hz)) - 1/2

Now, let's plug that into our funny little calculator:

n ≈ -42034.25

Oops! Looks like I made a clown-tastic mistake. It seems like I forgot to switch the numerator and denominator. Let me fix that for you:

n = (1.0 eV / (4.1357 x 10^-15 eV·s × 1.7 x 10^13 Hz)) - 1/2
n ≈ 1.45466 x 10^7

So, the approximate value of n for a state having an energy of 1.0 eV is approximately 1.45466 x 10^7. But hey, don't take my word for it, I'm just a clown-bot after all!

To calculate the difference in energy between allowed oscillator states, we can use the formula:

ΔE = hf

where ΔE is the difference in energy, h is Planck's constant (6.626 × 10^-34 J·s), and f is the frequency of oscillation.

(a) Converting the frequency to the SI unit of Hz:

f = 1.7 × 10^13 Hz

Using the formula, we can calculate the difference in energy:

ΔE = (6.626 × 10^-34 J·s) × (1.7 × 10^13 Hz)

Now, since the question asks for the answer in eV (electron volts), we need to convert the energy from joules to eV.

1 eV is equal to 1.6 × 10^-19 J. So, dividing the energy by the conversion factor:

ΔE = (6.626 × 10^-34 J·s) × (1.7 × 10^13 Hz) / (1.6 × 10^-19 J/eV)

Simplifying the expression:

ΔE ≈ 11.2 eV

Therefore, the difference in energy between allowed oscillator states is approximately 11.2 eV.

(b) Now, let's find the approximate value of n for a state having an energy of 1.0 eV.

We can use the formula for the energy of an oscillator state:

E = (n + 1/2) * hf

Solving for n:

n = (E / hf) - 1/2

Given that:

E = 1.0 eV
f = 1.7 × 10^13 Hz
h = 6.626 × 10^-34 J·s

Converting the energy from eV to joules:

E = 1.0 eV × (1.6 × 10^-19 J/eV)

Substituting the values into the equation:

n = [(1.0 eV × (1.6 × 10^-19 J/eV)) / ((6.626 × 10^-34 J·s) × (1.7 × 10^13 Hz))] - 1/2

Simplifying the expression:

n ≈ 341

Therefore, the approximate value of n for a state with an energy of 1.0 eV is 341.

To find the difference in energy between allowed oscillator states in a LiBr molecule, we'll use the formula:

ΔE = hf

Where:
ΔE is the difference in energy between states,
h is the Planck's constant (6.626 × 10^-34 Joule seconds or 4.136 × 10^-15 eV seconds),
and f is the frequency of oscillation.

(a) Let's plug in the values:

ΔE = (6.626 × 10^-34 eV s) × (1.7 × 10^13 Hz)
≈ 11.25 × 10^-21 eV

Therefore, the difference in energy between allowed oscillator states is approximately 11.25 × 10^-21 eV.

(b) Now, let's find the approximate value of n for a state with an energy of 1.0 eV. For an oscillator, the energy of a state is given by the formula:

E = (n + 1/2) × h × f

Rearranging the formula:

n = (E / (h × f)) - 1/2

Let's plug in the values:

n = (1.0 eV) / ((4.136 × 10^-15 eV s) × (1.7 × 10^13 Hz)) - 1/2
≈ 0.6

Therefore, the approximate value of n for a state with an energy of 1.0 eV is approximately 0.6.