Rotation Spectra - Basic transition question

[unscientific]

Homework Statement

In rotations of diatomic molecules such as HCL, the hamiltonian is found to be:

\hat H = \frac{\hat L^2}{2\mu a^2}

where $\mu$ is the reduced mass, a is the separation.

(a) Find the energy levels and separation.

(b) Explain why rotational spectra of HCl due to emission or absorption of electric dipole radiation consists of equally spaced lines. Given spacing $20.57cm^{-1}$, calculate the separation $a$.

Homework Equations

The Attempt at a Solution

Part(a)

E = \frac{l(l+1)\hbar^2}{2\mu a^2}

\Delta E = E_l - E_{l-1} = \frac{l(l+1) - l(l-1)}{2\mu a^2}\hbar^2 = \frac{l\hbar^2}{\mu a^2}

Part (b)

By dipole transition selection rules, $\Delta l = \pm 1$.

m not sure why it would be equally spaced, since spacing depends on $l$. I tried letting $l \rightarrow \infty$, which implies spacing goes to zero as $l(l+1) − l(l−1) \approx l^2−l^2 = 0$.

 

[DrClaude]

m not sure why it would be equally spaced, since spacing depends on $l$.

Imagine that $\Delta E$ was independent of $l$, how many spectral lines would be present in the rotational spectrum?

Start plotting a rotational spectrum for the $0 \rightarrow 1$, $1 \rightarrow 2$, $2 \rightarrow 3$, etc. You will see clearly the pattern.

 

[unscientific]

Imagine that $\Delta E$ was independent of $l$, how many spectral lines would be present in the rotational spectrum?

Start plotting a rotational spectrum for the $0 \rightarrow 1$, $1 \rightarrow 2$, $2 \rightarrow 3$, etc. You will see clearly the pattern.

The problem is, I can't show why $\Delta E$ is independent of $l$. If it is, then of course for each spontaneous emission it would be of equal energy.

 

[king vitamin]

The problem is, I can't show why $\Delta E$ is independent of $l$. If it is, then of course for each spontaneous emission it would be of equal energy.

This statement is false.

I think you need to keep thinking about your result. For example, what does the energy spectrum of your system look like? As a function of l? Maybe try plotting this (as suggested by DrClaude)?

 

[DrClaude]

The problem is, I can't show why $\Delta E$ is independent of $l$.

Your equation is correct, $\Delta E$ is dependent on $l$.

My guess is that you are stumbling here because you are not making the right connection between the energy levels and the spectrum.

 

[unscientific]

Imagine that $\Delta E$ was independent of $l$, how many spectral lines would be present in the rotational spectrum?

Start plotting a rotational spectrum for the $0 \rightarrow 1$, $1 \rightarrow 2$, $2 \rightarrow 3$, etc. You will see clearly the pattern.

From 1 to 0, I get $\frac{\hbar^2}{\mu a^2} = E_0$.

From 2 to 1, I get $2E_0$

From 3 to 2, I get $3E_0$

and so on..

As you can see here, the energy emitted gets bigger and bigger.

But the difference between one emission and the another is constant, $E_0$.

Does this mean that the spacing is constant?

Using $\frac{\hbar^2}{\mu a^2} = 20.57cm^{-1} = 2057 m^{-1}$, where $\mu \approx m_e$

I get $a = 4 \times 10^{-21} m$, which is ridulous.

 

[DrClaude]

But the difference between one emission and the another is constant, $E_0$.

Does this mean that the spacing is constant?

That indeed means that the spacing is constant. Each line in the spectrum is separated from the next by $E_0$ (think photon energy).

Using $\frac{\hbar^2}{\mu a^2} = 20.57cm^{-1} = 2057 m^{-1}$, where $\mu \approx m_e$

You have to use the reduced mass of HCl. Also, you will have to convert 20.57 cm^-1, which is wavenumbers, into energy.

 

[unscientific]

That indeed means that the spacing is constant. Each line in the spectrum is separated from the next by $E_0$ (think photon energy).


You have to use the reduced mass of HCl. Also, you will have to convert 20.57 cm^-1, which is wavenumbers, into energy.

Energy of emission is $\hbar k_l c$.

Thus, $\frac{\hbar^2}{\mu a^2} = \hbar c (\Delta k)$. Solving, I get $a = 3.24 \times 10^{-10} m$.

 

[DrClaude]

Energy of emission is $\hbar k_l c$.

In spectroscopy, "wavenumbers" mean $\bar{\nu} = 1 / \lambda$, so the energy of emission is actually $h c \bar{\nu}$.

Thus, $\frac{\hbar^2}{\mu a^2} = \hbar c (\Delta k)$. Solving, I get $a = 3.24 \times 10^{-10} m$.

Apart from having used $\hbar$ instead of $h$, as noted above, your answer is two orders of magnitude too small.

 

[unscientific]

In spectroscopy...

Ok, so it should be:

\frac{\hbar^2}{\mu a^2} = h c (\Delta \bar \nu)

Using that, and $\Delta \bar \nu = 2057$, I get: $1.28 \times 10^{-10} m$

Why is it missing two orders of magnitude? I used $\mu = \frac{35m_p^2}{35m_p + m_p} \approx m_p$

 

[DrClaude]

Ok, so it should be:

\frac{\hbar^2}{\mu a^2} = h c (\Delta \bar \nu)

For your information, $\bar\nu$ corresponds to the wavenumber of the photon. Therefore, you have $| \Delta E_{\mathrm{atom}} | = h c \bar{\nu}$, so there is no $\Delta \bar \nu$. (I put absolute values on $\Delta E_{\mathrm{atom}}$ because it can be negative or positive for emission or absorption.)

I used $\mu = \frac{35m_p^2}{35m_p + m_p} \approx m_p$

That's not good enough for spectroscopic work. Normally, the problem should specify which isotope of Cl it is, as you can easily distinguish ³⁵Cl from ³⁷Cl in a rotational spectrum. In the absence of this information, I think you should either choose one isotope or use the average atomic weigth, but you can't use something like $35 m_p$ or the approximation of neglecting the presence of H (you are not comparing a proton with an electron). Look up the exact values (Wolfram Alpha is a good resource).

Why is it missing two orders of magnitude?

I don't know. Maybe if you list all the constants you are using I can spot the error.

 

[unscientific]

For your information, $\bar\nu$ corresponds to the wavenumber of the photon. Therefore, you have $| \Delta E_{\mathrm{atom}} | = h c \bar{\nu}$, so there is no $\Delta \bar \nu$. (I put absolute values on $\Delta E_{\mathrm{atom}}$ because it can be negative or positive for emission or absorption.)That's not good enough for spectroscopic work. Normally, the problem should specify which isotope of Cl it is, as you can easily distinguish ³⁵Cl from ³⁷Cl in a rotational spectrum. In the absence of this information, I think you should either choose one isotope or use the average atomic weigth, but you can't use something like $35 m_p$ or the approximation of neglecting the presence of H (you are not comparing a proton with an electron). Look up the exact values (Wolfram Alpha is a good resource).I don't know. Maybe if you list all the constants you are using I can spot the error.

I was asked to use the mass of the isotope 35Cl. Now with $35m_p$ in 35Cl, $m_p$ in H, the reduced mass becomes $\frac{35m_p^2}{36m_p} \approx m_p$.

I use $\Delta \bar \nu = \nu_l - \nu_{l-1}$ to represent the spacing between emissions.