Question

what transition in the hydrogen spectrum would have the same wavelength as

the balmer transition n = 4 to n= 2 of He +spectra​

Answer 1

Answer:

4+2=5 same wavelength

Answer 2

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