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JEE Physics 2026 Question with Solution

The smallest wavelength of Lyman series is 91 nm91\ \text{nm}. The difference between the largest wavelengths of Paschen and Balmer series is nearly _____ nm\text{nm}.

  • A

    17841784

  • B

    12171217

  • C

    18751875

  • D

    15501550

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The smallest wavelength of Lyman series is λmin⁡=91 nm\lambda_{\min}=91\ \text{nm}.

Find: The difference between the largest wavelengths of Paschen and Balmer series.

For hydrogen spectrum, the smallest wavelength of Lyman series is

λmin⁡=1R=91 nm\lambda_{\min}=\frac{1}{R}=91\ \text{nm}

So,

1R=91 nm\frac{1}{R}=91\ \text{nm}

The largest wavelength in a spectral series corresponds to transition between nearest energy levels.

Largest wavelength of Balmer series corresponds to transition n=3→2n=3 \to 2:

λB=1R(122−132)\lambda_B=\frac{1}{R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)} λB=365R\lambda_B=\frac{36}{5R}

Largest wavelength of Paschen series corresponds to transition n=4→3n=4 \to 3:

λP=1R(132−142)\lambda_P=\frac{1}{R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)} λP=1447R\lambda_P=\frac{144}{7R}

Now, the required difference is

λP−λB=1447R−365R\lambda_P-\lambda_B=\frac{144}{7R}-\frac{36}{5R} λP−λB=46835R\lambda_P-\lambda_B=\frac{468}{35R}

Using 1R=91 nm\frac{1}{R}=91\ \text{nm},

λP−λB=46835×91\lambda_P-\lambda_B=\frac{468}{35}\times 91 λP−λB≈1217 nm\lambda_P-\lambda_B\approx 1217\ \text{nm}

Therefore, the difference is 1217 nm1217\ \text{nm} and the correct option is B.

Nearest-Level Transition Idea

Given: The smallest wavelength of Lyman series is 1R=91 nm\frac{1}{R}=91\ \text{nm}.

Find: The difference between the largest wavelengths of Paschen and Balmer series.

Use the fact that the largest wavelength in any series comes from the nearest higher level falling to that series level.

So directly take:

  • Balmer: 3→23 \to 2
  • Paschen: 4→34 \to 3

Then,

λB=1R(14−19)=365R\lambda_B=\frac{1}{R\left(\frac{1}{4}-\frac{1}{9}\right)}=\frac{36}{5R} λP=1R(19−116)=1447R\lambda_P=\frac{1}{R\left(\frac{1}{9}-\frac{1}{16}\right)}=\frac{144}{7R}

Hence,

λP−λB=1447R−365R=46835R\lambda_P-\lambda_B=\frac{144}{7R}-\frac{36}{5R}=\frac{468}{35R}

Now substitute 1R=91 nm\frac{1}{R}=91\ \text{nm}:

46835×91≈1217 nm\frac{468}{35}\times 91\approx 1217\ \text{nm}

Therefore, the correct option is B.

Common mistakes

  • Assuming the largest wavelength corresponds to the largest energy gap is incorrect. Since E=hcλE=\frac{hc}{\lambda}, the largest wavelength corresponds to the smallest energy gap, so use transitions between nearest levels.

  • Using the wrong transitions for Balmer and Paschen series is a common error. For the largest wavelength, use 3→23\to 2 for Balmer and 4→34\to 3 for Paschen, not transitions from very high nn.

  • Confusing the given smallest Lyman wavelength with a Balmer or Paschen wavelength is incorrect. The relation from the solution is λmin⁡=1R=91 nm\lambda_{\min}=\frac{1}{R}=91\ \text{nm}, which is used only to replace 1R\frac{1}{R} in the final expression.

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