Answer :
Final answer:
The wavelength of radiation to excite an electron in Li2+ from the first to the third Bohr orbit is calculated using the adapted Rydberg formula for hydrogen-like ions, considering the atomic number and applying the specific values for lithium's energy levels.
Explanation:
The wavelength of the radiation required to excite the electron in a doubly ionized lithium atom (Li2+) from the first to the third Bohr orbit can be derived using the Bohr model formula for the energy levels of a hydrogen-like atom:
\( E_n = -\frac{Z^2 R_H}{n^2} \)
Where \(E_n\) is the energy of the nth level, \(Z\) is the atomic number (for lithium \(Z=3\)), and \(R_H\) is the Rydberg constant for hydrogen. The energy of a photon emitted or absorbed during a transition is equal to the difference in energy levels:
\( \Delta E = E_1 - E_3 \)
To find the wavelength \( \lambda \), we use:
\( \lambda = \frac{hc}{\Delta E} \)
Here, \(h\) is Planck's constant, and \(c\) is the speed of light. After calculating the change in energy levels and converting it into the wavelength, we can determine which of the given options corresponds to the required wavelength. The detailed calculation will employ the Rydberg formula adapted for a hydrogen-like ion:
\( \frac{1}{\lambda} = Z^2 R \left( \frac{1}{n_1^2} - \frac{1}{n_3^2} \right) \)
With \(n_1 = 1\) and \(n_3 = 3\) for the first and third orbits respectively. Solving the above equation will yield the wavelength in meters, which can be converted to angstroms (Å) to match the given options.
