Question

The shortest wavelength of the Braqkett series of a hydrogen like atom (atomic number = Z) is the same as the shortest wavelength of the Balmar series of hydrogen atom. The value of Z is –

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (A)

2

Answer 2

Shortest wavelength of Bracket series corresponds to the transition of electron n₁ = 4 and n₂ = ¥ and the shortest wavelength of Balmer series corresponds to the transition of electron between n₁ = 2 and n₂ = ¥. So

(Z²) $\left( \frac{13.6}{16} \right)$ = $\left( \frac{13.6}{4} \right)$

\ Z² = 4 or Z = 2