Question

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

Answer 1

Rydberg equation is a mathematical relationship used to determine the wavelength of light emitted when an electron makes a transition from 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 $109677.57{\text{ c}}{{\text{m}}^{ - 1}}$.

Complete step by step 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 the Balmer transition $n = 4{\text{ to }}n = 2$ of ${\text{H}}{{\text{e}}^ + }$ 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 $n = 4{\text{ to }}n = 2$ of ${\text{H}}{{\text{e}}^ + }$ spectrum
Hence, equation (1) = equation (2)

\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 $$2 \to 1$$ of Lyman series in hydrogen spectrum have the same wavelength as Balmer transition $$n = 4{\text{ to }}n = 2$$ of $${\text{H}}{{\text{e}}^ + }$$ 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 helium atom.