Lowest Possible Sum of Charges for Hydrogen Like Species Transitions

For a hydrogen-like species, the wavelength $\lambda$ of light absorbed during an electronic transition from a lower energy level $n_1$ to a higher energy level $n_2$ is given by the Rydberg formula:

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

where $R_H$ is the Rydberg constant and $Z$ is the atomic number of the species.

Since $\mathbf{X}^{a+}$ and $\mathbf{Y}^{b+}$ are hydrogen-like species, each contains exactly $1$ electron. Thus, their charges $a$ and $b$ are related to their respective atomic numbers $Z_X$ and $Z_Y$ by: $a = Z_X - 1 \implies Z_X = a + 1$ $b = Z_Y - 1 \implies Z_Y = b + 1$

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Step 1: Transition in $\mathbf{X}^{a+}$

The transition occurs between $n = 1$ and $n = 2$, and the absorbed wavelength is $\lambda$:

$\frac{1}{\lambda} = R_H Z_X^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H Z_X^2 \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H Z_X^2$

$\implies \lambda = \frac{4}{3 R_H Z_X^2} \quad \text{--- (1)}$

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Step 2: Transition in $\mathbf{Y}^{b+}$

The transition occurs between $n = 2$ and $n = 4$, and the absorbed wavelength is $9\lambda$:

$\frac{1}{9\lambda} = R_H Z_Y^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H Z_Y^2 \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16} R_H Z_Y^2$

$\implies 9\lambda = \frac{16}{3 R_H Z_Y^2} \implies \lambda = \frac{16}{27 R_H Z_Y^2} \quad \text{--- (2)}$

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Step 3: Relating $Z_X$ and $Z_Y$

Equating equations (1) and (2):

$\frac{4}{3 R_H Z_X^2} = \frac{16}{27 R_H Z_Y^2}$

$\frac{1}{Z_X^2} = \frac{4}{9 Z_Y^2}$

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

Substituting $Z_X = a + 1$ and $Z_Y = b + 1$:

$2(a + 1) = 3(b + 1)$ $2a + 2 = 3b + 3 \implies 2a = 3b + 1$

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Step 4: Finding the lowest possible value of $(a + b)$

Since $a$ and $b$ are non-negative integers representing ionic charges:

1. For $b = 0 \implies 2a = 1 \implies a = 0.5$ (not an integer).
2. For $b = 1 \implies 2a = 4 \implies a = 2$ (valid integer solution).

With $a = 2$ and $b = 1$:

• $Z_X = 2 + 1 = 3$ (corresponding to $\text{Li}^{2+}$)
• $Z_Y = 1 + 1 = 2$ (corresponding to $\text{He}^+$)

Thus, the lowest possible value of $(a + b)$ is: $a + b = 2 + 1 = 3$