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Step-by-Step Solution
Step 1: Recognize the Hydrogen-like Atoms
All the given species—Hydrogen (${}_{1}H^{1}$), Deuterium (${}_{1}H^{2}$), singly ionized Helium (${}_{2}He^{4+}$), and doubly ionized Lithium (${}_{3}Li^{6++}$)—have only one electron. Such species are called hydrogen-like atoms, and they obey similar principles for electronic transitions.
Step 2: Recall the Relation for Wavelength in a Hydrogen-like Atom
For a hydrogen-like atom, the wave number $ \frac{1}{\lambda} $ of the photon emitted when the electron transitions from an initial energy level $n$ to a lower energy level $n_{1}$ is given by the Rydberg formula:
$ \displaystyle \frac{1}{\lambda} \;=\; R\,Z^{2}\,\Bigl[\;\frac{1}{n_{1}^{2}} \;-\;\frac{1}{n^{2}}\Bigr] $
Here:
$ R $ is the Rydberg constant.
$ Z $ is the atomic number (number of protons in the nucleus).
$ n $ is the higher energy level (initial), and $ n_{1} $ is the lower energy level (final).
Step 3: Apply the Formula to the $n=2$ to $n=1$ Transition
In the question, the transition is from $n=2$ to $n=1$, so the term $ \bigl[\tfrac{1}{n_{1}^{2}} - \tfrac{1}{n^{2}}\bigr] $ is common to all species because $n_{1} = 1$ and $n = 2$. Therefore, the wave number (and thus the wavelength) depends primarily on $Z^{2}$.
Hence,
$ \displaystyle \frac{1}{\lambda} \;\propto\; Z^{2} \quad\Longrightarrow\quad \lambda \;\propto\; \frac{1}{Z^{2}}. $
Step 4: Compare Atomic Numbers and Deduce the Wavelength Ratio
For the given ions/atoms:
Hydrogen (${}_{1}H^{1}$) has $Z = 1.$
Deuterium (${}_{1}H^{2}$) also has $Z = 1$ (same atomic number as Hydrogen).
Singly ionized Helium (${}_{2}He^{4+}$) has $Z = 2.$
Doubly ionized Lithium (${}_{3}Li^{6++}$) has $Z = 3.$
Since $ \lambda \propto \frac{1}{Z^{2}}, $ we have:
For $Z = 1$ (Hydrogen or Deuterium): $ \lambda \propto \frac{1}{1^{2}} = 1. $
For $Z = 2$ (He$^{+}$): $ \lambda \propto \frac{1}{2^{2}} = \frac{1}{4}. $
For $Z = 3$ (Li$^{2+}$): $ \lambda \propto \frac{1}{3^{2}} = \frac{1}{9}. $
Step 5: Express the Final Wavelength Ratios
Let $ \lambda_{1} $ be the wavelength for Hydrogen, $ \lambda_{2} $ for Deuterium, $ \lambda_{3} $ for singly ionized Helium, and $ \lambda_{4} $ for doubly ionized Lithium. Since Hydrogen and Deuterium both have $Z=1$, they have the same wavelength:
$ \lambda_{1} = \lambda_{2}. $
Helium ($Z=2$) has wavelength $ \frac{1}{4} $ of that for Hydrogen (or Deuterium):
$ \lambda_{3} = \frac{1}{4}\,\lambda_{1}. $
Lithium ($Z=3$) has wavelength $ \frac{1}{9} $ of that for Hydrogen (or Deuterium):
$ \lambda_{4} = \frac{1}{9}\,\lambda_{1}. $
Rewriting,
$ \lambda_{1} = \lambda_{2} = 4\,\lambda_{3} = 9\,\lambda_{4}. $
This matches the correct answer:
$ \displaystyle \lambda_{1} = \lambda_{2} = 4\,\lambda_{3} = 9\,\lambda_{4}. $
Final Answer
The correct relationship among the wavelengths is
$ \lambda_{1} = \lambda_{2} = 4\,\lambda_{3} = 9\,\lambda_{4}. $
