Question

What transition in the hydrogen spectrum would have the same wavelength as the balmer transition n 4?

Answer 1

Answer:

The wave number for Balmer transition of n =4 to n=2 of He⁺ spectrum would be:-

\overline V = RZ^{2} \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

\overline V = \dfrac{1}{\lambda} = R(2)^{2} \left( \dfrac{1}{4} - \dfrac{1}{6} \right)

\overline V = \dfrac{1}{\lambda} = 4R \left( \dfrac{4-1}{16} \right)

\overline V = \dfrac{1}{\lambda} = \dfrac{3R}{4}

\lambda = \dfrac{4}{3R}

Acc. to question, the transition for Hydrogen will have the same wavelength:-

\implies R(1)^{2}\left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right) = \dfrac{3R}{4}

\left[ \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right] = \dfrac{3}{4}

This equation can only be true when n_1 = 1 and n_2 = 2

\therefore The transition of electron from n = 2 to n = 1 in Hydrogen would have the same wavelength as transition of electron from n = 4 to n = 1 in He^{+}

Answer 2

Answer:

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

Explanation: