$\text{X}^{a+}$ and $\text{Y}^{b+}$ are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers $n = 1$ and $n = 2$ of $\text{X}^{a+}$ is $\lambda$. The wavelength of light absorbed during the transition between the states with principal quantum numbers $n = 2$ and $n = 4$ of $\text{Y}^{b+}$ is $9\lambda$. The lowest possible value of $(a + b)$ is ____.

For a hydrogen-like species, the wavelength of the absorbed light during a transition from $n_1$ to $n_2$ is given by the Rydberg formula:

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

For $\text{X}^{a+}$, the transition is from $n = 1$ to $n = 2$:

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

For $\text{Y}^{b+}$, the transition is from $n = 2$ to $n = 4$:

$\dfrac{1}{9\lambda} = R Z_Y^2 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right) = R Z_Y^2 \left( \dfrac{1}{4} - \dfrac{1}{16} \right) = \dfrac{3}{16} R Z_Y^2 \Rightarrow 9\lambda = \dfrac{16}{3 R Z_Y^2}$

Substituting the expression for $\lambda$ from the first equation into the second:

$9 \left( \dfrac{4}{3 R Z_X^2} \right) = \dfrac{16}{3 R Z_Y^2}$

$\dfrac{12}{Z_X^2} = \dfrac{16}{3 Z_Y^2}$

$36 Z_Y^2 = 16 Z_X^2$

$9 Z_Y^2 = 4 Z_X^2 \Rightarrow 3 Z_Y = 2 Z_X$

$\dfrac{Z_X}{Z_Y} = \dfrac{3}{2}$

For the lowest possible integer values of atomic numbers, we get $Z_X = 3$ and $Z_Y = 2$.

Since $\text{X}^{a+}$ and $\text{Y}^{b+}$ are hydrogen-like species, they must contain exactly $1$ electron. Therefore, the charge on the ion is $(Z - 1)$.

For $\text{X}^{a+}$ ($Z_X = 3$): $a = 3 - 1 = 2$

For $\text{Y}^{b+}$ ($Z_Y = 2$): $b = 2 - 1 = 1$

The lowest possible value of $(a + b)$ is $2 + 1 = 3$.

Answer: $3$