A spectral line of Lyman series of H-atom has a frequency of 2.466 x 10^15 s^-1. What is the transition responsible for this spectral line ? (R = 1.096 x 10^7 m^-1)

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q121 – AP-EAMCET 2025

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Home Chemistry Structure of AtomQuestion #597

A spectral line of Lyman series of H-atom has a frequency of

2.466 \times 10^{15} s^{- 1}

. What is the transition responsible for this spectral line ? (

R = 1.096 \times 10^{7} m^{- 1}

)

Clue 1 of 3: Hydrogen Spectrum Series

Recall the different spectral series for the hydrogen atom and the corresponding final energy levels ( $n_{1}$ ). The Lyman series specifically involves transitions to the ground state, meaning $n_{1} = 1$ .

🧠 Full Solution Path

Step 1: Identify Given Values and Series Type✦ Active

The problem provides the frequency of a spectral line in the Lyman series of a hydrogen atom and the Rydberg constant. We need to find the electron transition responsible for this line.

Given values:

ν = 2.466 \times 10^{15} s^{- 1}

R = 1.096 \times 10^{7} m^{- 1}

Speed of light, c = 3 \times 10^{8} m/s

For the Lyman series, the electron transitions to the ground state, which means the final principal quantum number is $n_{1} = 1$ .

💡 Teacher's Secret Hint

Remember that each spectral series (Lyman, Balmer, Paschen, etc.) corresponds to a specific final energy level ( $n_{1}$ ). Lyman series always ends at $n_{1} = 1$ .

Step 2: Apply the Rydberg Formula for Frequency○ Expand

The frequency of a spectral line for a hydrogen atom is given by the Rydberg formula:

ν = R c (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

Substitute the known values for $ν$ , $R$ , $c$ , and $n_{1} = 1$ into the formula:

2.466 \times 10^{15} = (1.096 \times 10^{7}) \times (3 \times 10^{8}) (\frac{1}{1^{2}} - \frac{1}{n_{2}^{2}})

💡 Teacher's Secret Hint

Ensure consistent units. The Rydberg constant is in $m^{- 1}$ and speed of light in $m/s$ , so their product $R c$ will be in $s^{- 1}$ , matching the frequency unit.

Step 3: Solve for

n_{2}

○ Expand

First, calculate the product $R c$ :

R c = (1.096 \times 10^{7} m^{- 1}) \times (3 \times 10^{8} m/s) = 3.288 \times 10^{15} s^{- 1}

Now, substitute this back into the equation from Step 2:

2.466 \times 10^{15} = 3.288 \times 10^{15} (1 - \frac{1}{n_{2}^{2}})

Divide both sides by $3.288 \times 10^{15}$ :

\frac{2.466 \times 10^{15}}{3.288 \times 10^{15}} = 1 - \frac{1}{n_{2}^{2}}

0.750 = 1 - \frac{1}{n_{2}^{2}}

Rearrange to solve for $\frac{1}{n_{2}^{2}}$ :

\frac{1}{n_{2}^{2}} = 1 - 0.750 = 0.250

Now, solve for $n_{2}^{2}$ and then $n_{2}$ :

n_{2}^{2} = \frac{1}{0.250} = 4

n_{2} = \sqrt{4} = 2

Thus, the transition is from $n_{2} = 2$ to $n_{1} = 1$ .

💡 Teacher's Secret Hint

Be careful with algebraic manipulation, especially when isolating $n_{2}^{2}$ . Double-check your calculations to avoid errors.

Step 4: Match with Options○ Expand

The calculated transition is from $n_{2} = 2$ to $n_{1} = 1$ . Comparing this with the given options, Option 1 matches this result.

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Chemistry Question 121 – AP-EAMCET 2025

Clue 1 of 3: Hydrogen Spectrum Series

🧠 Full Solution Path