In Li$$^{++}$$, electron in first Bohr orbit is excited to a level by a radiation of wavelength $$\lambda$$. When the ion gets de-excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $$\lambda$$?
(Given: h = $$6.63 \times 10^{-34}$$ J s; c = $$3 \times 10^8$$ m s$$^{-1}$$)

We are dealing with the hydrogen-like ion Li$$^{++}$$, for which the nuclear charge is $$Z = 3$$ and only one electron revolves. An electron that was initially in the ground state $$n = 1$$ absorbs a photon of wavelength $$\lambda$$ and jumps to some higher level $$n = n_f$$. After that, the ion returns to the ground state by emitting photons in every possible way, and the statement tells us that exactly six distinct spectral lines are recorded.

For any one-electron atom or ion, the total number of different lines obtainable when an electron falls from a level $$n_f$$ back to the ground level is given by the well-known combinatorial result

$$\text{Number of lines} \; = \; \frac{n_f(n_f-1)}{2}.$$

We are told this number is $$6$$, so we write

$$\frac{n_f(n_f-1)}{2}=6.$$

Multiplying both sides by $$2$$ gives

$$n_f(n_f-1)=12.$$

Expanding the left side, we obtain the quadratic equation

$$n_f^2 - n_f - 12 = 0.$$

Factoring,

$$(n_f-4)(n_f+3)=0,$$

which furnishes the physical root

$$n_f = 4.$$

Hence the electron was excited from $$n = 1$$ to $$n = 4$$ by absorbing the photon of wavelength $$\lambda$$.

Next we evaluate the energy difference between these two levels. For a hydrogen-like species the Bohr energy formula is first recalled:

$$E_n = -\dfrac{13.6\,\text{eV}\, Z^2}{n^2}.$$

Substituting $$Z = 3$$ for Li$$^{++}$$, we get

$$E_n = -\dfrac{13.6 \times 3^2}{n^2}\;\text{eV} = -\dfrac{13.6 \times 9}{n^2}\;\text{eV}.$$

So, for the ground state $$n=1$$,

$$E_1 = -\dfrac{13.6 \times 9}{1^2} = -122.4 \;\text{eV},$$

and for the excited state $$n=4$$,

$$E_4 = -\dfrac{13.6 \times 9}{4^2} = -\dfrac{122.4}{16} = -7.65 \;\text{eV}.$$

The photon had to supply the positive energy difference

$$\Delta E = E_4 - E_1 = (-7.65\,\text{eV}) - (-122.4\,\text{eV}) = 114.75 \;\text{eV}.$$

We convert this energy to joules using $$1\;\text{eV} = 1.602 \times 10^{-19}\;\text{J}:$$

$$\Delta E = 114.75 \times 1.602 \times 10^{-19}\;\text{J} = 1.838295 \times 10^{-17}\;\text{J}.$$

The photon wavelength is connected to its energy via the Einstein relation $$E = h c / \lambda.$$ Stating this formula explicitly,

$$\lambda = \dfrac{h\,c}{\Delta E}.$$

With the given constants $$h = 6.63 \times 10^{-34}\;\text{J s}$$ and $$c = 3.00 \times 10^{8}\;\text{m s}^{-1},$$ we have

$$h\,c = (6.63 \times 10^{-34})(3.00 \times 10^{8}) = 1.989 \times 10^{-25}\;\text{J m}.$$

Dividing by $$\Delta E$$,

$$\lambda = \dfrac{1.989 \times 10^{-25}}{1.838295 \times 10^{-17}}\;\text{m}.$$

Carrying out the division,

$$\lambda = 1.0819 \times 10^{-8}\;\text{m} = 10.819 \times 10^{-9}\;\text{m} = 10.8\;\text{nm}.$$

Among the given choices, $$10.8\;\text{nm}$$ corresponds to Option A.

Hence, the correct answer is Option A.