JEE Advanced 2026 — Paper 2 — CHEM-III solution

JEE Advanced2026Paper 2CHEM-III

$X^{a +}$ and $Y^{b +}$ 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 $X^{a +}$ is $λ$ . The wavelength of light absorbed during the transition between the states with principal quantum numbers $n = 2$ and $n = 4$ of $Y^{b +}$ is $9 λ$ . The lowest possible value of $(a + b)$ is _____.

Your answer

Solution

Rydberg formula for a hydrogenic ion. $\frac{1}{λ} = R_{H} z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$ .

For X $^{a +}$ ( $n = 1 \to 2$ ): $\frac{1}{λ} = R_{H} z_{X}^{2} (1 - \frac{1}{4}) = \frac{3}{4} R_{H} z_{X}^{2}$ .

For Y $^{b +}$ ( $n = 2 \to 4$ ): $\frac{1}{9 λ} = R_{H} z_{Y}^{2} (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} R_{H} z_{Y}^{2}$ .

Divide. $9 = \frac{z _{X}^{2}}{z _{Y}^{2}} \cdot \frac{3/4}{3/16} = 4 \cdot \frac{z _{X}^{2}}{z _{Y}^{2}} \Rightarrow \frac{z _{X}^{2}}{z _{Y}^{2}} = \frac{9}{4} \Rightarrow \frac{z _{X}}{z _{Y}} = \frac{3}{2}$ .

Smallest integer Z pair: $z_{X} = 3$ (Li, so X $^{a +}$ = Li $^{2 +}$ , $a = 2$ ), $z_{Y} = 2$ (He, so Y $^{b +}$ = He $^{+}$ , $b = 1$ ).

$∴ (a + b)_{m i n} = 2 + 1 = 3$ .

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