Electron transitions for the Hydrogen atom n=7 n=6 n=5 n=4 n=3 Paschen series E(n) to E(n=3) n=2 Balmer series Ein) to E(n=2) n=1 Lyman series E(n) to E(n=1) The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from n = ∞ to n = 2. What would be the specific wavelength of such a line? [Hint :1/λ = RH(1/n² - 1/n;²), R₁ being Rydberg constant = 1.097 x 107 /m] 560 nm 365 nm 400 nm 600 nm Brackett series E(n) to E(n=4)

icon
Related questions
Question
### Electron Transitions for the Hydrogen Atom

The diagram shows the electron transitions in a hydrogen atom for different series. Each colored line represents electron transitions from higher energy levels (n) to a specific lower energy level.

- **Lyman Series (Ultraviolet Region):**
  - Transitions from n ≥ 2 to n = 1
  - Represented by orange lines.

- **Balmer Series (Visible Region):**
  - Transitions from n ≥ 3 to n = 2
  - Represented by green lines.

- **Paschen Series (Infrared Region):**
  - Transitions from n ≥ 4 to n = 3
  - Represented by teal lines.

- **Brackett Series (Infrared Region):**
  - Transitions from n ≥ 5 to n = 4
  - Represented by blue lines.

### Problem Statement

The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from \( n = \infty \) to \( n = 2 \). What would be the specific wavelength of such a line?

### Hint

\[
\frac{1}{\lambda} = R_H\left(\frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2}\right)
\]
where \( R_H \) (Rydberg constant) = \( 1.097 \times 10^7 \, \text{m}^{-1} \).

### Calculation

For series limit of the Balmer series:
- \( n_{\text{initial}} = \infty \)
- \( n_{\text{final}} = 2 \)

\[
\frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = R_H\left(\frac{1}{4} - 0 \right) = \frac{R_H}{4}
\]

\[
\lambda = \frac{4}{R_H} = \frac{4}{1.097 \times 10^7 \, \text{m}^{-1}}
\]

\[
\lambda \approx 364.6 \, \text{nm}
\]

### Multiple Choice Question

Select the correct specific wavelength for the Balmer series limit.
Transcribed Image Text:### Electron Transitions for the Hydrogen Atom The diagram shows the electron transitions in a hydrogen atom for different series. Each colored line represents electron transitions from higher energy levels (n) to a specific lower energy level. - **Lyman Series (Ultraviolet Region):** - Transitions from n ≥ 2 to n = 1 - Represented by orange lines. - **Balmer Series (Visible Region):** - Transitions from n ≥ 3 to n = 2 - Represented by green lines. - **Paschen Series (Infrared Region):** - Transitions from n ≥ 4 to n = 3 - Represented by teal lines. - **Brackett Series (Infrared Region):** - Transitions from n ≥ 5 to n = 4 - Represented by blue lines. ### Problem Statement The series limit wavelength of the Balmer series is emitted as the electron in the hydrogen atom falls from \( n = \infty \) to \( n = 2 \). What would be the specific wavelength of such a line? ### Hint \[ \frac{1}{\lambda} = R_H\left(\frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2}\right) \] where \( R_H \) (Rydberg constant) = \( 1.097 \times 10^7 \, \text{m}^{-1} \). ### Calculation For series limit of the Balmer series: - \( n_{\text{initial}} = \infty \) - \( n_{\text{final}} = 2 \) \[ \frac{1}{\lambda} = R_H\left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = R_H\left(\frac{1}{4} - 0 \right) = \frac{R_H}{4} \] \[ \lambda = \frac{4}{R_H} = \frac{4}{1.097 \times 10^7 \, \text{m}^{-1}} \] \[ \lambda \approx 364.6 \, \text{nm} \] ### Multiple Choice Question Select the correct specific wavelength for the Balmer series limit.
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

Blurred answer