Question

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number 'Z'. The value of 'Z' is

• A. \(6\)
• B. \(4\)
• C. \(3\)
• D. \(2\)

Hint:

The shortest wavelength of a series corresponds to the transition from infinity.

Answer 1

The Correct Option is D

Step 1: Scaling Idea:
Energy levels of a hydrogen-like atom scale as $Z^2/n^2$. The limit of a series ending at level $n_1$ has energy $\propto Z^2/n_1^2$.

Step 2: Match:
Balmer limit of hydrogen: $1/4$. Brackett limit of the ion: $Z^2/16$. Setting $Z^2/16=1/4$ gives $Z^2=4$.

Step 3: Answer:
$Z=2$, which is He$^+$. Option (D).

Final Answer:
Option (D). \[ \boxed{\text{(D) } 2} \]