A gas of hydrogen-like ions is prepared in a specific excited state...

Let the principal quantum number of the excited state $A$ be $n$.

The total number of different wavelengths emitted is $1 + 5 = 6$.

The number of spectral lines emitted when a gas de-excites from state $n$ to the ground state is given by $\dfrac{n(n-1)}{2}$.

Equating this to $6$:

$\dfrac{n(n-1)}{2} = 6$

$n^2 - n - 12 = 0$

$(n-4)(n+3) = 0$

Since $n$ must be a positive integer, $n = 4$.

The wavelength of the first line of the Lyman series for hydrogen ($Z=1$) corresponds to the transition from $n=2$ to $n=1$. Its wavelength $\lambda$ is given by:

$\dfrac{1}{\lambda} = R (1)^2 \left( \dfrac{1}{1^2} - \dfrac{1}{2^2} \right) = \dfrac{3R}{4}$

For a hydrogen-like ion of atomic number $Z$, the wavelength of an emitted photon during a transition from $n_2$ to $n_1$ (where $n_2 \le 4$) is:

\dfrac{1}{\lambda} = R Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

Given that this wavelength is equal to the first line of the Lyman series of hydrogen:

R Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) = \dfrac{3R}{4}

Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) = \dfrac{3}{4}

Checking for $Z = 2$ ($\text{He}^+$):

4 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) = \dfrac{3}{4}

\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} = \dfrac{3}{16}

This equation is satisfied for $n_1 = 2$ and $n_2 = 4$, which are valid states since the gas is excited to $n = 4$.

Thus, the gas is $\text{He}^+$ and the principal quantum number of state $A$ is $4$.