X^{a+} and Y^{b+} are hydrogen-like species. The wavelength of light absorbed during the transition

Q: Why do we use $Z^2$ in the Rydberg formula?

The Rydberg formula for hydrogen assumes $Z=1$. For hydrogen-like species (ions with only one electron like $\text{He}^+$, $\text{Li}^{2+}$), the increased nuclear charge pulls the electron tighter, scaling the energy levels by a factor of $Z^2$.

Q: Why did we divide the two equations instead of substituting?

Division is a powerful algebraic tool in physics and chemistry to eliminate unknown constants. Here, dividing instantly removes $R_H$ and $\lambda$, leaving a clean equation with only our target variables, $Z_x$ and $Z_y$.

Q: How did we know $Z_x=3$ and $Z_y=2$ are the exact values?

The math only gave us the ratio $Z_x/Z_y = 3/2$. However, the question specifically asked for the **lowest possible value** of $(a+b)$. This constraint forces us to pick the smallest possible integers that fit the ratio, which are 3 and 2.

Q: What does 'hydrogen-like species' mean practically?

It means the atom or ion has exactly **one electron** orbiting the nucleus, just like hydrogen. This is crucial because the Bohr model and the simple Rydberg formula only work perfectly for single-electron systems.

Q: How did we calculate the charges $a$ and $b$ from $Z$?

The number of electrons in an ion is the atomic number ($Z$) minus its charge. Since hydrogen-like species have 1 electron, we use the formula: $1 = Z - \text{charge}$. So, charge $= Z - 1$.

Question

$	ext{X}^{a+}$ and $	ext{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 $	ext{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 $	ext{Y}^{b+}$ is $9\lambda$. The lowest possible value of $(a+b)$ is _____.

Sigma Percentile · Accepted Answer

### The Quantum Setup

Imagine you are peering into the quantum world of two mysterious ions, $	ext{X}^{a+}$ and $	ext{Y}^{b+}$. We are told they are "hydrogen-like," which is a massive clue. It means that despite having heavier nuclei than hydrogen, they have been stripped of all their electrons except for one lonely survivor.

For species $	ext{X}^{a+}$, this single electron absorbs a photon of wavelength $\lambda$ and leaps from the ground state ($n=1$) to the first excited state ($n=2$). Meanwhile, in species $	ext{Y}^{b+}$, the electron absorbs a much lower energy photon with a wavelength of $9\lambda$, causing it to jump from $n=2$ to $n=4$. Our mission is to decode the identities of these ions and find the lowest possible sum of their charges, $(a+b)$.

### The Master Equation: Rydberg Formula

To connect the absorbed wavelength, the atomic number ($Z$), and the energy levels, we deploy our trusty tool: the **Rydberg formula**.

Because these are hydrogen-like species and not just hydrogen, we must include the $Z^2$ term. The increased nuclear charge pulls the electron tighter, scaling the energy levels dramatically. The formula is:

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

Let's apply this to species $	ext{X}^{a+}$. The transition is from $n_1=1$ to $n_2=2$. Substituting these values, we get:

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

Now, let's do the exact same thing for species $	ext{Y}^{b+}$. Here, the absorbed wavelength is $9\lambda$, and the transition is from $n_1=2$ to $n_2=4$:

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

### The Mathematical Duel

We now have two beautiful equations. Our goal is to find the relationship between the atomic numbers $Z_x$ and $Z_y$. The most elegant way to eliminate the constants, $R_H$ and $\lambda$, is to simply divide the first equation by the second equation.

$$ \frac{\frac{1}{\lambda}}{\frac{1}{9\lambda}} = \frac{R_H Z_x^2 \left( \frac{3}{4} ight)}{R_H Z_y^2 \left( \frac{3}{16} ight)} $$

On the left side, the fraction flips and gives us exactly $9$. On the right side, the Rydberg constant $R_H$ cancels out. The fraction $\frac{3}{4}$ divided by $\frac{3}{16}$ simplifies perfectly to $4$.

$$ 9 = \frac{Z_x^2}{Z_y^2} 	imes 4 $$

Rearranging the terms, we get:

$$ \frac{Z_x^2}{Z_y^2} = \frac{9}{4} $$

Taking the square root of both sides, we find that the ratio of the atomic numbers is exactly $3:2$.

$$ \frac{Z_x}{Z_y} = \frac{3}{2} $$

### Decoding "Hydrogen-Like"

The question specifically asks for the **lowest possible value** of $(a+b)$. This implies we need the lowest possible integer values for the atomic numbers. The smallest positive integers that satisfy a $3:2$ ratio are simply $Z_x = 3$ and $Z_y = 2$.

Now, here is the catch. What does "hydrogen-like" actually mean practically? It means the species has exactly **one electron**.

The number of electrons in an ion is calculated as its atomic number minus its charge.

For species $	ext{X}^{a+}$ with atomic number $3$ (which is Lithium), to have only one electron, it must have lost two electrons.

$$ 3 - a = 1 \implies a = 2 \quad (	ext{This is } 	ext{Li}^{2+}) $$

For species $	ext{Y}^{b+}$ with atomic number $2$ (Helium), it must have lost one electron.

$$ 2 - b = 1 \implies b = 1 \quad (	ext{This is } 	ext{He}^{+}) $$

### The Final Tally

Finally, we just need to add the charges $a$ and $b$ together to find our target value.

$$ a + b = 2 + 1 = 3 $$

And that is our final, lowest possible value! Always pay close attention to constraints like "lowest possible" in JEE Advanced problems, as they are the key to locking down the final numerical answer.

Enter Numerical Value:

Visualized Solution

$Visualizing the Transitions$

$The Rydberg Formula$

$Equation for Species X$

$Equation for Species Y$

$Dividing the Equations$

$Simplifying the Ratio$

$Finding the Ratio of Z$

$Applying the Constraint$

$Decoding Hydrogen-Like Species$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Bohr's Model

The Quantum Setup

The Master Equation

The Mathematical Duel

Decoding "Hydrogen-Like"

The Final Tally

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Transitions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Rydberg Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for Species X

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation for Species Y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dividing the Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Ratio of Z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Constraint

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decoding Hydrogen-Like Species

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward