1

λ

=

−

v

= 1.097 × 107Z2[

1

n12

−

1

n22

]m−1
where, n1 = number of the lower energy level, n2 = number of the higher energy level, Z = atomic number, λ = wavelength,

−

v

= wavenumber
For He+,

1

λ

= 1.097 × 107 × (2)2[

1

(2)2

−

1

(4)2

]m−1
= 1.097 × 107×4(

3

16

) = 1.097 × 107×

3

4

For H atom,

1

λ

= 1.097 × 107×(1)2×

3

4

= 1.097 × 107[

1

nL2

−

1

nH2

]
[

1

nL2

−

1

nH2

] =

3

4

. Thig gives nL = 1 , nH = 2
The transition nL = 1 to nH = 2 in H-atom would have the same wavelength as Balmer transition n = 4 to n = 2 of He+.