# Homework Help: Revolutions per second for hydrogen molecule

1. Oct 4, 2014

### Effect

1. The problem statement, all variables and given/known data
The distance between the both nucleons in the H2 molecule is 0.0741 nm
The energy for the 3 lowest radiation levels of the molecule are:
$E_{0}=0 \\ E_{1}=0.0151 eV \\ E_{2}=0.0453 eV$
How many revolutions per second does the molecule rotate with in the different states?

2. Relevant equations
See attachment

3. The attempt at a solution
How many revolutions per second does the molecule rotate with? I thought that if f in the last equation above is frequency then the rotational speed would be omega=f*2pi. So if we rewrite the last equation we get
$f=\frac{E_{v}}{(\nu+1/2)h} \Leftrightarrow \omega=\frac{2\pi* E_{v}}{(\nu+1/2)h}$

Of course the first level gives zero rev/s but for the others I get the wrong answer. When putting in the above values, I get

$\omega=\frac{2\pi* 0.0151* 1.606*10^{-19}}{(1+1/2)*6.626*10^{-34}}=1.53*10^{14} rev/s$ which is incorrect. Also the other value gives the wrong answer in this equation.

$\omega_0 = 0 \\ \omega_1 = 5.2*10^{12} \\ \omega_2 = 9.0*10^{12} \\ (rev/s)$

I don't understand how to get to these answers. Shouldn't the above equation be used? Or is it something else that needs to be taken into consideration? Please help how to get to these answers.

#### Attached Files:

• ###### rotation.png
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Last edited: Oct 4, 2014
2. Oct 4, 2014

### vela

Staff Emeritus
The equation you started with applies to vibrations of the molecule, not rotations. It doesn't apply here.

You need to go back and review rotational kinematics to find out how the energy is related to the frequency of rotation.

3. Oct 4, 2014

### Effect

Do you mean rotational dynamics $$E_{rotational}=1/2 I\omega^2$$
This?

I still don't get the right answer.

4. Oct 4, 2014

### vela

Staff Emeritus
Yes, that's right. It worked for me. Show your work.

5. Oct 4, 2014

### Effect

$E_{rotational}=1/2 I\omega^2 \Leftrightarrow \omega=\sqrt{\frac{2E}{I}} \\ I=1/2mr^2$
$$\Rightarrow \omega=\sqrt{\frac{4E}{mr^2}}$$
$\Rightarrow \omega=\sqrt{\frac{4*0.0151*1.606*10^{-19}}{1.66*10^{-27}*(7.41*10^{-11})^2}}=1.03*10^{13}$

Last edited: Oct 4, 2014
6. Oct 4, 2014

### vela

Staff Emeritus
Divide by $2\pi$ since you want the result in rev/s.

7. Oct 4, 2014

### Effect

$\frac{1.03*10^{13}}{2\pi}=1.64*10^{12}$
Still not correct, correct answer should be $5.2*10^{12}$

8. Oct 4, 2014

### vela

Staff Emeritus
I don't get $1.03 \times 10^{13}$ in the numerator from the expression you gave above. I get $3.26 \times 10^{13}$.

9. Oct 4, 2014

### Effect

You are correct, I must have made a careless mistake somewhere. Thank you for your assistance!

10. Oct 5, 2014

### ehild

Under the square root, the result is 1.0642x1027. You took the square root of 1.0642, and added the magnitude 1013 instead of the square root of 10.642 x 10 26.