Chemistry Q121 – AP-EAMCET 2026

Home Chemistry Structure of AtomQuestion #595

In an atomic spectrum of hydrogen, a series of lines with wavelengths at 656.46, 486.27,

x

and 410.29 nm was obtained. What is the value of

x

(in nm)? (

R_{H} = 1.097 \times 10^{7} m^{- 1}

)

Clue 1 of 3: The Rydberg Formula for Hydrogen

Understand that atomic spectra result from electron transitions between quantized energy levels. For hydrogen, these transitions are described by the Rydberg formula.

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Step 1: Identify the spectral series and energy levels✦ Active

The given wavelengths 656.46 nm, 486.27 nm, and 410.29 nm are characteristic of the visible region of the hydrogen spectrum, which corresponds to the Balmer series. In the Balmer series, electrons transition to the principal quantum number $n_{1} = 2$ . The wavelengths correspond to transitions from higher energy levels ( $n_{2}$ ) to $n_{1} = 2$ . The wavelengths decrease as $n_{2}$ increases.

The sequence of lines in the Balmer series starting from the longest wavelength (lowest energy difference) is:

- $n_{2} = 3 \to n_{1} = 2$ (H-alpha): $λ = 656.46 nm$

- $n_{2} = 4 \to n_{1} = 2$ (H-beta): $λ = 486.27 nm$

- $n_{2} = 5 \to n_{1} = 2$ (H-gamma): This is the unknown wavelength $x$ .

- $n_{2} = 6 \to n_{1} = 2$ (H-delta): $λ = 410.29 nm$

Therefore, the value of $x$ corresponds to the transition from $n_{2} = 5$ to $n_{1} = 2$ .

Step 2: Apply the Rydberg formula○ Expand

The Rydberg formula for the wavelength of light emitted from hydrogen-like atoms is:

\frac{1}{λ} = R_{H} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

where $R_{H}$ is the Rydberg constant, $λ$ is the wavelength, $n_{1}$ is the principal quantum number of the lower energy level, and $n_{2}$ is the principal quantum number of the higher energy level.

Given $R_{H} = 1.097 \times 10^{7} m^{- 1}$ .

For the unknown wavelength $x$ , we have $n_{1} = 2$ and $n_{2} = 5$ .

💡 Teacher's Secret Hint

Ensure all units are consistent. If $R_{H}$ is in $m^{- 1}$ , the calculated wavelength will be in meters.

Step 3: Calculate the wavelength○ Expand

Substitute the values of $R_{H}$ , $n_{1}$ , and $n_{2}$ into the Rydberg formula:

\frac{1}{x} = 1.097 \times 10^{7} m^{- 1} (\frac{1}{2^{2}} - \frac{1}{5^{2}})

\frac{1}{x} = 1.097 \times 10^{7} m^{- 1} (\frac{1}{4} - \frac{1}{25})

Find a common denominator for the fractions:

\frac{1}{x} = 1.097 \times 10^{7} m^{- 1} (\frac{25 - 4}{100})

\frac{1}{x} = 1.097 \times 10^{7} m^{- 1} (\frac{21}{100})

Now, calculate the value of $x$ :

x = \frac{100}{21 \times 1.097 \times 10^{7} m^{- 1}}

x = \frac{100}{23.037 \times 10^{7} m^{- 1}}

x = 4.3408 \times 10^{- 7} m

Step 4: Convert the wavelength to nanometers○ Expand

Since $1 m = 10^{9} nm$ , convert the calculated wavelength from meters to nanometers:

x = 4.3408 \times 10^{- 7} m \times (\frac{10^{9} nm}{1 m})

x = 434.08 nm

Comparing this value with the given options, $434.17 nm$ is the closest.

💡 Teacher's Secret Hint

Always pay attention to the requested units in the final answer. Often, wavelengths are asked in nanometers (nm) or angstroms (Å).

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Chemistry Question 121 – AP-EAMCET 2026

Clue 1 of 3: The Rydberg Formula for Hydrogen

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