\( X^{a+} \) 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 \( \lambda \). The wavelength of light absorbed during the transition between the states with principal quantum numbers \( n = 2 \) and \( n = 4 \) of \( Y^{b+} \) is \( 9\lambda \). The lowest possible value of \( (a + b) \) is _________.
Detailed Step-by-Step Solution
This is an excellent multi-step problem that combines the Rydberg formula with the definition of hydrogen-like species.
Step 1: Set up equations using the Rydberg Formula
The Rydberg formula is given by: \( \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)
For species \( X^{a+} \) (Let atomic number be \( Z_x \)):
Transition is \( n=1 \) to \( n=2 \) with wavelength \( \lambda \).
\( \frac{1}{\lambda} = R_H Z_x^2 \left( \frac{3}{4} \right) \quad \text{--- (Equation 1)} \)
For species \( Y^{b+} \) (Let atomic number be \( Z_y \)):
Transition is \( n=2 \) to \( n=4 \) with wavelength \( 9\lambda \).
\( \frac{1}{9\lambda} = R_H Z_y^2 \left( \frac{3}{16} \right) \quad \text{--- (Equation 2)} \)
Step 2: Find the ratio of atomic numbers (\(Z_x\) and \(Z_y\))
Divide Equation 1 by Equation 2:
\( 9 = \frac{Z_x^2}{Z_y^2} \times \left( \frac{3}{4} \times \frac{16}{3} \right) \)
\( 9 = \frac{Z_x^2}{Z_y^2} \times 4 \)
Rearrange to solve for the ratio of \(Z\):
Since \( Z_x \) and \( Z_y \) are atomic numbers, they must be integers. The lowest possible integer values that satisfy this ratio are \( \mathbf{Z_x = 3} \) and \( \mathbf{Z_y = 2} \).
Step 3: Relate Atomic Number to Charge (\(a\) and \(b\))
The problem states \( X^{a+} \) and \( Y^{b+} \) are hydrogen-like species. A hydrogen-like species has exactly 1 electron.
The formula for the number of electrons is: \( \text{Electrons} = Z - \text{Charge} \).
- For \( X^{a+} \) (\(Z_x = 3\)):
\( a = 3 - 1 = \mathbf{2} \) (This is \( \text{Li}^{2+} \)). - For \( Y^{b+} \) (\(Z_y = 2\)):
\( b = 2 - 1 = \mathbf{1} \) (This is \( \text{He}^+ \)).
Conclusion: We need to find the lowest possible value of \( (a + b) \).
\( a + b = 2 + 1 = \mathbf{3} \).
Mastering JEE Advanced Logic
What makes this a JEE Advanced caliber question is the combination of mathematical ratios with a fundamental chemical definition. Many students successfully calculate \(Z_x/Z_y = 3/2\) but fail to connect that hydrogen-like means the species has only 1 electron, meaning the charge \(a\) must be \(Z_x - 1\).
To master these interconnected concepts and tackle multiple-concept numericals with confidence, be sure to review our comprehensive guide on the Structure of Atom Class 11 Chemistry.
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