Question

what transition in hydrogen spectrum will have same lambda as Balmer transition from
4 orbit ----> 2 orbit of helium spectrum.​

Answer 1

For an atom = 1/λ = RHZ2 (1/n12-1/n22)

For He+ spectrum Z = 2, n2=4, n1= 2

Therefore = 1/λ = RH x 4(1/22-1/42) = 3RH/4

For hydrogen spectrum = 3RH/4, Z = 1

Therefore = 1/λ = RH X 1(1/n12-1/n22)

Or

RH (1/n12-1/n22) = 3RH/4

Or

1/n12-1/n22 = 3/4

Which can be so for n1=1 & n2 = 2, i.e the transition is from n = 2 to n=1