Question 10

Using the Rydberg formula for hydrogen-like species: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ For $X^{a+}$, $Z_X = a+1$. Transition $n=1 \to n=2$: $\frac{1}{\lambda} = R (a+1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R (a+1)^2 \left( \frac{3}{4} \right) \dots (1)$ For $Y^{b+}$, $Z_Y = b+1$. Transition $n=2 \to n=4$: $\frac{1}{9\lambda} = R (b+1)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R (b+1)^2 \left( \frac{3}{16} \right) \dots (2)$ Dividing equation (1) by equation (2): $9 = \frac{(a+1)^2}{(b+1)^2} \times \frac{3/4}{3/16} = \frac{(a+1)^2}{(b+1)^2} \times 4$ $9 = 4 \left( \frac{a+1}{b+1} \right)^2 \implies \frac{3}{2} = \frac{a+1}{b+1}$ $3(b+1) = 2(a+1) \implies 3b + 3 = 2a + 2 \implies 2a - 3b = 1$ For the lowest possible values of $a$ and $b$ (which must be positive integers for ions): If $b=1$, then $2a - 3 = 1 \implies 2a = 4 \implies a = 2$. Thus, $a=2$ and $b=1$ are the smallest integers. $a+b = 2+1 = 3$.