Question

The shortest wavelength of H atom is the Lyman series is λ₁. The longest wavelength in the Balmer series of He⁺ is :- 

(1) 5λ₁/9

(2) 27λ₁/5

(3) 9λ₁/5

(4) 36λ₁/5

Answer 1

(3) 9λ₁/5

As we know ΔE = hc/λ

So λ = hc/ΔE for λ minimum i.e. 

Shortest; ΔE = maximum 

For Lyman series n = 1 & for ΔE_max 

Transition must be form n = ∞ to n = 1 

So

For longest wavelength ΔE = minimum for Balmer series n = 3 to n = 2 will have ΔE minimum

For He⁺ Z = 2 

So

Comments

Very good explanation