Question

What transition in hydrogen spectrum have the same wavelength as Balmer transition $n = 4\;{\rm{ to }}\;n = 2{\rm{\; of\;H}}{{\rm{e}}^ + }$ spectrum?

Answer 1

Rydberg equation is a mathematical relationship used to determine the wavelength of light emitted when an electron makes a transition form one energy level to another energy level. A photon of light is emitted when an electron moves from a higher energy level to lower energy level. Rydberg formula can be applied to the spectra of different elements.
The Rydberg’s constant is $109677.57{\rm{ c}}{{\rm{m}}^{ - 1}}$.

Complete answer:
For a transition ${n_2} \to {n_1}$ for an atom having atomic number Z, the wavelength is given by the following expression.

$\dfrac{1}{\lambda } = R{Z^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$

Here R is Rydberg constant and the above equation is called Rydberg equation.
The atomic number of hydrogen is 1 and that of helium is 2.

Consider ${n_2} \to {n_1}$ transitions in the hydrogen spectrum. The Rydberg equation will be
$\dfrac{1}{\lambda } = R{\left( 1 \right)^2}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$

$\dfrac{1}{\lambda } = R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ … …(1)

Consider ${n_4} \to {n_2}$ transitions in the hydrogen spectrum. The Rydberg equation will be
$\dfrac{1}{\lambda } = R{\left( 2 \right)^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right]$

$\dfrac{1}{\lambda } = 4R\left[ {\dfrac{1}{4} - \dfrac{1}{{16}}} \right]$

$\dfrac{1}{\lambda } = R\left[ {1 - \dfrac{1}{4}} \right]$

$\dfrac{1}{\lambda } = \dfrac{{3R}}{4}$ … …(2)

The transition in hydrogen spectrum has the same wavelength as Balmer transition $n = 4\;{\rm{ to }}\;n = 2{\rm{\;of\;H}}{{\rm{e}}^ + }$ spectrum
Hence, equation (1) = equation (2)
$R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{{3R}}{4}$
$\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{3}{4}$
For the first line in the Lyman series, ${n_2} \to {n_1}$ or $2 \to 1$.
$\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{3}{4}$
$\left[ {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right] = \dfrac{3}{4}$
$\left[ {1 - \dfrac{1}{4}} \right] = \dfrac{3}{4}$
$\dfrac{3}{4} = \dfrac{3}{4}$
Since both sides of the equation are equal, the transition is indeed the first line $2 \to 1$ in the Lyman series of the hydrogen spectrum.
Hence, the transition of Lyman series in hydrogen spectrum have the same wavelength as Balmer transition $n = 4\;{\rm{\; to }}\;n = 2\;{\rm{ of\;H}}{{\rm{e}}^ + }$ spectrum.

Note: In hydrogen spectrum, each transition is associated with a particular wavelength. Similarly, in the helium spectrum, each transition is associated with a particular wavelength. It is possible that the wavelength of a particular transition in the helium spectrum matches with the wavelength of a particular transition in the hydrogen atom. Properly chose the transition in the hydrogen atom that has the same wavelength as that of the given transition of the helium atom.