Structure of Atom

Concept:Use the Rydberg formula for hydrogen-like species to relate wavelength, atomic number, and principal quantum numbers.Explanation:For a hydrogen-like species, the wavelength of absorbed light during a transition is given by:$$\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$Here, $R_H$ is the Rydberg constant and $Z$ is the atomic number.For $\mathbf{X}^{a+}$, transition from $n=1$ to $n=2$ gives:$$\frac{1}{\lambda} = R_H Z_X^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H Z_X^2 \left( \frac{3}{4} \right) \quad \ldots (1)$$For $\mathbf{Y}^{b+}$, transition from $n=2$ to $n=4$ gives wavelength $9\lambda$:$$\frac{1}{9\lambda} = R_H Z_Y^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H Z_Y^2 \left( \frac{3}{16} \right) \quad \ldots (2)$$Divide equation (1) by (2):$$\frac{1/\lambda}{1/(9\lambda)} = \frac{Z_X^2 \cdot \frac{3}{4}}{Z_Y^2 \cdot \frac{3}{16}}$$This simplifies to:$$9 = \frac{Z_X^2}{Z_Y^2} \times 4$$Thus,$$\frac{Z_X^2}{Z_Y^2} = \frac{9}{4} \quad \Rightarrow \quad \frac{Z_X}{Z_Y} = \frac{3}{2}$$Take the smallest integer atomic numbers satisfying the ratio:$Z_X = 3$ and $Z_Y = 2$.Since $\mathbf{X}^{a+}$ is hydrogen-like, it has lost one electron, so atomic number $Z_X = a + 1$.Thus, $a = 3 - 1 = 2$.Similarly, $b = 2 - 1 = 1$.Therefore, the lowest possible value of $(a+b)$ is $2 + 1 = 3$.Answer:3