Find Wavelength of Emitted Photon in Helium Ion Transition

To find the wavelength of the emitted photon, we first determine the principal quantum numbers ($n_1$ and $n_2$) corresponding to the given radii of the orbits.

The radius of the $n$-th orbit for a hydrogen-like species ($\text{He}^+$, where atomic number $Z = 2$) is given by: $r_n = \frac{n^2}{Z} \times a$

Given:

• Bohr radius, $a = 52.9\text{ pm}$
• Initial radius, $r_2 = 105.8\text{ pm}$
• Final radius, $r_1 = 26.45\text{ pm}$

For the initial orbit ($r_2$): $105.8 = \frac{n_2^2}{2} \times 52.9$ $n_2^2 = \frac{105.8 \times 2}{52.9} = 4 \implies n_2 = 2$

For the final orbit ($r_1$): $26.45 = \frac{n_1^2}{2} \times 52.9$ $n_1^2 = \frac{26.45 \times 2}{52.9} = 1 \implies n_1 = 1$

Thus, the transition is from $n_2 = 2$ to $n_1 = 1$.

The energy of an electron in the $n$-th orbit is given by: $E_n = -R_H \frac{Z^2}{n^2}$

The energy of the emitted photon during the transition ($\Delta E$) is: $\Delta E = E_2 - E_1 = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Substituting the given values: $\Delta E = (2.2 \times 10^{-18}\text{ J}) \times (2)^2 \times \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$ $\Delta E = 2.2 \times 10^{-18} \times 4 \times \frac{3}{4} = 6.6 \times 10^{-18}\text{ J}$

The wavelength $\lambda$ of the emitted photon is related to $\Delta E$ by: $\Delta E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta E}$

Substituting the values for Planck's constant ($h$) and the speed of light ($c$): $\lambda = \frac{6.6 \times 10^{-34}\text{ J s} \times 3 \times 10^8\text{ m s}^{-1}}{6.6 \times 10^{-18}\text{ J}}$ $\lambda = 3 \times 10^{-8}\text{ m} = 30\text{ nm}$

Final Answer: The wavelength of the emitted photon is 30 nm.