Question

which transition in hydrogen atomic spectrum will have same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum ?​

Answer 1

Used formula:

1/λ=RZ²[1/n₁²-1/n₂²]

where Z=atomic number of atom

Balmer transition in He⁺ :

1/λ=4R[1/4-1/16]....(Z=2 for He)1

1/λ=3R/4

For H atom:

1/λ=R[1/n₁²-1/n₂²]

(Z=1 for H)

Putting n₁=1 and n₂=2

1/λ=3R/4

So the transition from 2→1 in H atom will give same wavelength as Balmer transition in He atom.