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

Question

$Li^{+2} \longrightarrow Li^{+3} + e^-$ :
Energy required for this process.
Given ionisation energy for ground state of hydrogen atom is $2.17 \times 10^{-18} \text{ J}$.

Answer 1

The Lyman, Balmer, and Paschen series are spectral line series related to the electron transitions in a hydrogen atom. In atomic physics, these series correspond to different electronic transitions, with Lyman series involving transitions to the ground state (n=1), Balmer series to the second energy level (n=2), and Paschen series to the third energy level (n=3). To solve this problem, we need to find the largest wavelength difference between the Balmer and Paschen series.

The formula for the wavelength of lines in any hydrogen spectral series is given by the Rydberg formula:

$\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

where:

$R$ is the Rydberg constant, approximately $1.097 \times 10^7 \ \text{m}^{-1}$.
$n_1$ is the lower energy level.
$n_2$ is the higher energy level.

For the Balmer series: $n_1 = 2$

The longest wavelength line in the Balmer series corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.

Using the formula, we can calculate:

$\frac{1}{\lambda_{\text{Balmer}}} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right)$

$\frac{1}{\lambda_{\text{Balmer}}} = R \left(\frac{1}{4} - \frac{1}{9}\right)$

$\frac{1}{\lambda_{\text{Balmer}}} = R \left(\frac{5}{36}\right)$

$\lambda_{\text{Balmer}} = \frac{36}{5R}$

For the Paschen series: $n_1 = 3$

The longest wavelength line in the Paschen series corresponds to the transition from $n_2 = 4$ to $n_1 = 3$.

Using the formula, we can calculate:

$\frac{1}{\lambda_{\text{Paschen}}} = R \left(\frac{1}{3^2} - \frac{1}{4^2}\right)$

$\frac{1}{\lambda_{\text{Paschen}}} = R \left(\frac{1}{9} - \frac{1}{16}\right)$

$\frac{1}{\lambda_{\text{Paschen}}} = R \left(\frac{7}{144}\right)$

$\lambda_{\text{Paschen}} = \frac{144}{7R}$

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

$\Delta \lambda = \lambda_{\text{Paschen}} - \lambda_{\text{Balmer}}$

Substituting the values we calculated:

$\Delta \lambda = \left(\frac{144}{7R}\right) - \left(\frac{36}{5R}\right)$

Simplifying the expression:

$\Delta \lambda = \frac{720 - 252}{35R} = \frac{468}{35R}$

Calculating with $R = 1.097 \times 10^7 \ \text{m}^{-1}$ and converting to nanometers:

$\Delta \lambda \approx 1217 \ \text{nm}$

The correct answer is, therefore, 1217 nm.

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

Show Hint

The Correct Option is B

Solution and Explanation

Top Questions on Atomic Physics

Questions Asked in JEE Main exam