- #1

Yroyathon

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## Homework Statement

What is the minimum (nonzero) rotational energy of the <sup>14</sup>N_2 molecule?

## Homework Equations

E_rot = L * (L + 1) * E_0 * (m_e) / (4*M)

M = (M_1 + M_2) / (M_1 * M_2)

## The Attempt at a Solution

first off, is E_0 = 13.6 eV? that's what it was in the similar problem on molecular hydrogen. If it isn't...well, how do I calculate it for this situation?

Ok, so the minimal non-zero E has to be when L=1, since L=0 yields E=0. With L=1, we get

E_rot = (E_0 * m_e) / (2 * M)

From the far-above equation for reduced mass M, I plug in 7 times the mass of a proton for each of the nitrogen nuclei, and get M = (7*m_p*7*m_p)/(7*m_p + 7*m_p). After simplifying, M = (7*m_p)/2

Plugging in E_0, m_e = .00091*10^(-27) kg, m_p = 1.672*10^(-27) kg, and M into the above equation for E_rot, I get

E_rot = (13.6 * .00091*10^(-27) * 2) / (2 * 7 * 1.672*10^(-27)), and after simplifying

E_rot = 0.00105742 eV. But this is wrong.

I thought I was pretty careful. Can anyone see a calculation error, or a mistake in my understanding of how to use these equations with this particular molecular nitrogen situation?

Many thanks.

,Yroyathon