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Step-by-Step Solution
Step 1: Identify the number of spectral lines and determine the excited state
• We are told that when the Li++ ion (which is hydrogen-like with atomic number Z = 3) de-excites back to the ground state, it emits a total of six distinct spectral lines.
• The formula for the number of lines coming from an upper level n2 down to n1 (including all possible intermediate transitions) is
$ \frac{n_2(n_2 - 1)}{2} $.
• Setting this equal to 6:
$ \frac{n_2(n_2 - 1)}{2} = 6 \implies n_2(n_2 - 1)=12 $.
• Solving, we find n2 = 4. Therefore, the electron must have been excited from the first orbit (n = 1) to the fourth orbit (n = 4).
Step 2: Use the Bohr model formula for a hydrogen-like ion
• For a hydrogen-like species of atomic number Z, the energy $E_n$ of the electron in the nth orbit is given by
$ E_n = -13.6 \times Z^2 \left(\frac{1}{n^2}\right) \text{ eV}. $
• The energy difference for a transition from n = 1 to n = 4 is:
$ \Delta E = E_4 - E_1. $
Step 3: Relate energy difference to radiation wavelength
• When an electron transitions from n = 1 to n = 4, the photon energy $ \Delta E $ is absorbed. This energy is related to the wavelength $ \lambda $ of the radiation by
$ \Delta E = \frac{hc}{\lambda}, $
where $h$ is Planck’s constant and $c$ is the speed of light.
• Alternatively, one can use the Rydberg formula for hydrogen-like species:
$ \frac{1}{\lambda} = R \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right), $
where $ R \approx 1.097 \times 10^7 \,\text{m}^{-1}, \, Z=3, \, n_1=1, \, n_2=4. $
Step 4: Calculate the wavelength using the Rydberg formula
Substitute $ Z=3, n_1=1, n_2=4 $ into the Rydberg expression:
\[
\frac{1}{\lambda}
= \bigl(1.097 \times 10^7 \,\text{m}^{-1}\bigr) \times 9
\left(\frac{1}{1^2} - \frac{1}{4^2}\right).
\]
\[
= 1.097 \times 10^7 \,\text{m}^{-1} \times 9
\left(1 - \frac{1}{16}\right).
\]
\[
= 9.873\times10^8 \left(\frac{15}{16}\right)
= 9.873\times10^8 \times 0.9375
\approx 9.256\times10^8 \,\text{m}^{-1}.
\]
Taking the reciprocal:
\[
\lambda = \frac{1}{9.256\times10^8 \,\text{m}^{-1}}
\approx 1.08 \times 10^{-9} \,\text{m}
= 10.8 \,\text{nm}.
\]
Step 5: Final answer
The wavelength $ \lambda $ required for the transition is therefore 10.8 nm.
