Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer 1

As per the hydrogen spectral series-The emission spectrum of atomic H is divided into some spectral series, with wavelengths given by the Rydberg formula.

The observed spectral lines are due to the electron making transitions between 2 energy levels in an atom.

For  transition in the hydrogen, the spectrum would have the same wavelength as the Balmer transition is He+,1λ=RHz2[122−142]

For H,1λ=RH[1n21−1n22]

Since λ is same

∴Z2[122−142]=[1n21−1n22]

z=2

[112−122]=[1n21−1n22]

n1=1 and n2=2