Lyman-Series: calculating the nuclear mass of an isotope

[LeoJakob]

Assuming I have a given wavelength λ of the hydrogen spectrum and I want to calculate the nuclear mass of the isotope, is my approach correct?

Rydberg-formula:

$$\frac{1}{\lambda} = R_M \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

with $$R_M = \frac{R_\infty}{1 + \frac{m_e}{M}}$$

Setting $n_1 = 1$ and $n_2 = 2$, we get:

$$\frac{1}{\lambda} = R_M \left( \frac{1}{1^2} - \frac{1}{2^2} \right)
\Leftrightarrow \frac{1}{\lambda} = R_M \left( 1 - \frac{1}{4} \right)
\Leftrightarrow \frac{1}{\lambda} = R_M \left( \frac{3}{4} \right)
\Leftrightarrow \frac{1}{\lambda} = \frac{3R_M}{4}
\Leftrightarrow R_M = \frac{4}{3\lambda}$$

Substituting $R_M$in the definition:

$$\frac{4}{3\lambda} = \frac{R_\infty}{1 + \frac{m_e}{M}}\Leftrightarrow 1 + \frac{m_e}{M} = \frac{3\lambda R_\infty}{4}\Leftrightarrow \frac{m_e}{M} = \frac{3\lambda R_\infty}{4} - 1 \Leftrightarrow M = \frac{m_e}{\frac{3\lambda R_\infty}{4} - 1}$$


Kind regards :)

 

[TSny]

Your calculations look correct to me.

However, your result for the nuclear mass $M$ will probably not be very accurate. The Rydberg formula is only approximate. A full quantum mechanical treatment of the hydrogen atom (including effects of relativity and electron spin) leads to a modification of your result that can be expressed as $$M = \frac{m_e}{\frac{3 \lambda R_{\infty}}{4}(1+\epsilon)-1}$$ where $\epsilon$ is a very small, dimensionless number that comes out of the long, messy calculation. $\epsilon$ is of the order of the square of the fine structure constant $\alpha$. So $\epsilon$ is of the order of $10^{-5}$.

At first, you might think that the very small value of $\epsilon$ would not make much difference in the calculation of $M$. However, the denominator in the expression for $M$ is of the form $x - 1$ where $x$ is a number very close to 1. So, small changes in $x$ can make a big difference in $M$.

 

[Mordred]

I'm glad you answered I didn't see anything wrong either but I kept second guessing myself so chose to let someone else reply. Glad to see that at least I wasn't incorrect in both the answer and the choice of waiting to see if I was correct.