If shortest wavelength of hydrogen atom in Lyman series is $$x$$, then longest wavelength in Balmer series of He$$^-$$ is:

1. Core Formula

The Rydberg formula for the wavelength of light emitted during electronic transitions is:

1 / λ = R × Z² × [ (1 / n₁²) - (1 / n₂²) ]

Where:

• λ = wavelength
• R = Rydberg constant
• Z = atomic number
• n₁ = lower energy level
• n₂ = higher energy level

2. Case 1: Shortest Wavelength of Hydrogen in Lyman Series

• For the Hydrogen atom, the atomic number (Z) = 1.
• For the Lyman series, the lower energy level (n₁) = 1.
• The shortest wavelength corresponds to the maximum energy transition, which occurs when an electron drops from infinity (n₂ = ∞).
• Given that this wavelength is x:

1 / x = R × 1² × [ (1 / 1²) - (1 / ∞) ]

1 / x = R × 1 × [ 1 - 0 ]

1 / x = R ⟶ Therefore, R = 1 / x

3. Case 2: Longest Wavelength of He⁺ in Balmer Series

• For the Helium ion (He⁺), the atomic number (Z) = 2.
• For the Balmer series, the lower energy level (n₁) = 2.
• The longest wavelength corresponds to the minimum energy transition, which occurs from the very next shell (n₂ = 3).

Let's substitute these values into the formula:

1 / λ_He⁺ = R × 2² × [ (1 / 2²) - (1 / 3²) ]

1 / λ_He⁺ = 4R × [ (1 / 4) - (1 / 9) ]

1 / λ_He⁺ = 4R × [ (9 - 4) / 36 ]

1 / λ_He⁺ = 4R × [ 5 / 36 ]

1 / λ_He⁺ = 5R / 9

4. Substitute R = 1 / x

1 / λ_He⁺ = (5 / 9) × (1 / x)

1 / λ_He⁺ = 5 / 9x

Taking the reciprocal to solve for λ_He⁺:

λ_He⁺ = 9x / 5

Correct Answer:

9x / 5