Question:
The atomic number (Z) of a hydrogen like atom whose shortest wavelength of Brackett Series is same as the shortest wavelength of Balmer series of hydrogen atom is
The atomic number (Z) of a hydrogen like atom whose shortest wavelength of Brackett Series is same as the shortest wavelength of Balmer series of hydrogen atom is
Show Hint
Shortest wavelength always means $n_{2} = \infty$. The equation simplifies directly to balancing $\frac{Z^2}{n_{1}^2}$ values between the two configurations.
Updated On: Jun 3, 2026
- Z = 1
- Z = 2
- Z = 3
- Z = 4
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The Correct Option is B
Solution and Explanation
Step 1: Concept
The wave number for a hydrogen-like atom is given by Rydberg's formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, where $R$ is the Rydberg constant and $Z$ is the atomic number.
Step 2: Meaning
The shortest wavelength (series limit) occurs when an electron transitions from infinity ($n_2 = \infty$) to a lower energy level ($n_1$). For the Balmer series, $n_1 = 2$. For the Brackett series, $n_1 = 4$.
Step 3: Analysis
For the Balmer series of hydrogen ($Z=1$): $$\frac{1}{\lambda_{\text{Balmer}}} = R (1)^2 \left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4}$$ For the Brackett series of the hydrogen-like ion: $$\frac{1}{\lambda_{\text{Brackett}}} = R Z^2 \left(\frac{1}{4^2} - \frac{1}{\infty}\right) = \frac{R Z^2}{16}$$ Equating the wavelengths as given in the problem: $$\frac{R}{4} = \frac{R Z^2}{16} \implies 4 = Z^2 \implies Z = 2$$
Step 4: Conclusion
The atomic number $Z$ of the atom must be 2.
Final Answer: (B)
The wave number for a hydrogen-like atom is given by Rydberg's formula: $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, where $R$ is the Rydberg constant and $Z$ is the atomic number.
Step 2: Meaning
The shortest wavelength (series limit) occurs when an electron transitions from infinity ($n_2 = \infty$) to a lower energy level ($n_1$). For the Balmer series, $n_1 = 2$. For the Brackett series, $n_1 = 4$.
Step 3: Analysis
For the Balmer series of hydrogen ($Z=1$): $$\frac{1}{\lambda_{\text{Balmer}}} = R (1)^2 \left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4}$$ For the Brackett series of the hydrogen-like ion: $$\frac{1}{\lambda_{\text{Brackett}}} = R Z^2 \left(\frac{1}{4^2} - \frac{1}{\infty}\right) = \frac{R Z^2}{16}$$ Equating the wavelengths as given in the problem: $$\frac{R}{4} = \frac{R Z^2}{16} \implies 4 = Z^2 \implies Z = 2$$
Step 4: Conclusion
The atomic number $Z$ of the atom must be 2.
Final Answer: (B)
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