The wave number of the limiting line of the series (visible) in hydrogen spectrum is:
27, 419 cm-1
The question asks about the wave number of the limiting line in the visible series of the hydrogen spectrum. Let's break down the key terms:
The wave number ($\bar{\nu}$) of a spectral line in the hydrogen spectrum is given by the Rydberg formula:
\(\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\)
Where:
For the limiting line of the visible series (Balmer series):
Substitute these values into the Rydberg formula:
\(\bar{\nu} = 109677 \text{ cm}^{-1} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)\)
\(\bar{\nu} = 109677 \text{ cm}^{-1} \left( \frac{1}{4} - 0 \right)\)
\(\bar{\nu} = 109677 \text{ cm}^{-1} \times \frac{1}{4}\)
Now, perform the calculation:
\(\bar{\nu} = \frac{109677}{4} \text{ cm}^{-1}\)
\(\bar{\nu} \approx 27419.25 \text{ cm}^{-1}\)
The calculated wave number is approximately \(27419.25 \text{ cm}^{-1}\). Looking at the given options, the closest value is \(27,419 \text{ cm}^{-1}\).
| Series | Lower Level ($n_1$) | Upper Level ($n_2$) | Region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4, ... , &(\infty) | Ultraviolet |
| Balmer (Visible) | 2 | 3, 4, 5, ... , &(\infty) | Visible |
| Paschen | 3 | 4, 5, 6, ... , &(\infty) | Infrared |
| Brackett | 4 | 5, 6, 7, ... , &(\infty) | Infrared |
| Pfund | 5 | 6, 7, 8, ... , &(\infty) | Infrared |
Thus, the wave number for the limiting line of the visible series (Balmer series, $n_1=2$, $n_2=\infty$) in the hydrogen spectrum is approximately \(27,419 \text{ cm}^{-1}\).
Understanding the different series and their corresponding transitions is crucial for solving problems related to the hydrogen spectrum.
The energy levels of a hydrogen atom are quantized, meaning they can only exist at specific discrete values. The energy of the \(n^{th}\) level is given by \(E_n = -\frac{13.6}{n^2}\) eV. When an electron moves from a higher energy level \(n_2\) to a lower energy level \(n_1\), it emits a photon with energy \(E = E_{n_2} - E_{n_1}\). This energy is related to the wave number by the equation \(E = hc\bar{\nu}\), where \(h\) is Planck's constant and \(c\) is the speed of light. The Rydberg formula is derived directly from these energy level transitions.
Each series has a shortest wavelength (highest wave number) and a longest wavelength (lowest wave number).
The visible series (Balmer) is particularly important because it was the first series to be discovered and studied in detail due to its wavelengths falling within the visible part of the spectrum.
The correct order of electromagnetic spectrum with decreasing frequency is:
The value of the proportionality constant μo/(4π) is equal to _________ Tm/A.
Match List I with List II
List – I | List – II | ||
US New Military Bands for Microwaves | Frequency range in GHz | ||
A. | H band | I. | 2.000 ‐ 3.000 GHz |
B. | J band | II. | 4.000 ‐ 6.000 GHz |
C. | G band | III. | 6.000 ‐ 8.000 GHz |
D. | E band | IV. | 10.000 ‐ 20.000 GHz |
Choose the correct answer from the options given below:
The wavelength in the bright-line emission spectrum of an element are