Question

Orbital energies in single-electron atoms or ions, such as $\mathrm{He}^{+}$, can be described with an equation similar to the Balmer-Rydberg equation:$\frac{1}{\lambda}=Z^2 R_{\infty}\left[\frac{1}{m^2}-\frac{1}{n^2}\right]$where $Z$ is the atomic number. What wavelength of light in nm is emitted when the electron in $\mathrm{He}^{+}$falls from $n=3$ to $n=2$ ?

Solution

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\begin{align*} &\frac{1}{\lambda} = \text{Z}^{2} \cdot \text{R}_\infty \cdot \text{c} \cdot [\frac{1}{m^2} - \frac{1}{n^2}]& \end{align*}

$\lambda$ - wavelength

$\text{R}_\infty$ - Rydberg constant equal to $1.097 \cdot 10^{-2}\ \text{nm}^{-1}$

c - speed of light, value is $3.00 \cdot 10^{8}$ m/s:

Z - Atomic number

m, n - $\text{m and n represent integers with n $>$ m}$.

$\text{\underline{\text{This is the Balmer-Ridberg equation}}}$