- #1
LeoJakob
- 22
- 1
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 :)
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 :)