Advertisements
Advertisements
प्रश्न
An electron in a hydrogen atom makes transitions from orbits of higher energies to orbits of lower energies.
- When will such transitions result in (a) Lyman (b) Balmer series?
- Find the ratio of the longest wavelength in the Lyman series to the shortest wavelength in the Balmer series.
Advertisements
उत्तर
(i) Emission spectrum of Hydrogen atom: Lyman and Balmer series
- Lyman Series: When electrons will jump from a higher energy orbit to n = 1 orbit.
- Balmer Series: When electrons will jump from a higher energy orbit yo n = 2 orbits.
(ii) Longest wavelength of Lyman series:
`1/λ_"L" = "R"_"H"[1/1^2 - 1/2^2] = "R"_"H" [3/4]`
Shortest wavelength of Balmer series:
`1/λ_"B" = "R"_"H"[1/2^2 - 1/(∞^2)] = "R"_"H" [1/4]`
Now,
`λ_"L"/λ_"B" = [4/3]/[4/1] = 1/3`
APPEARS IN
संबंधित प्रश्न
When is Hα line in the emission spectrum of hydrogen atom obtained? Calculate the frequency of the photon emitted during this transition.
Calculate the minimum wavelength of the spectral line present in Balmer series of hydrogen
The ground state energy of hydrogen atom is −13.6 eV. If and electron make a transition from the energy level −0.85 eV to −3.4 eV, calculate spectrum does his wavelength belong?
In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory
Hydrogen spectrum consists of discrete bright lines in a dark background and it is specifically known as hydrogen emission spectrum. There is one more type of hydrogen spectrum that exists where we get dark lines on the bright background, it is known as the absorption spectrum. Balmer found an empirical formula by the observation of a small part of this spectrum and it is represented by
`1/lambda = "R"(1/2^2 - 1/"n"^2)`, where n = 3, 4, 5,....
For Lyman series, the emission is from first state to nth state, for Paschen series, it is from third state to nth state, for Brackett series, it is from fourth state to nth state and for P fund series, it is from fifth state to nth state.
Which series of hydrogen spectrum corresponds to ultraviolet region?
Hydrogen spectrum consists of discrete bright lines in a dark background and it is specifically known as hydrogen emission spectrum. There is one more type of hydrogen spectrum that exists where we get dark lines on the bright background, it is known as absorption spectrum. Balmer found an empirical formula by the observation of a small part of this spectrum and it is represented by
`1/lambda = "R"(1/2^2 - 1/"n"^2)`, where n = 3, 4, 5,....
For Lyman series, the emission is from first state to nth state, for Paschen series, it is from third state to nth state, for Brackett series, it is from fourth state to nth state and for P fund series, it is from fifth state to nth state.
Rydberg constant is ______.
In the following atoms and molecules for the transition from n = 2 to n = 1, the spectral line of minimum wavelength will be produced by
The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 Å. Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum.
Name the series of lines of hydrogen spectrum which lie in the visible region.
Match List I with List II.
| List-I (Spectral Lines of Hydrogen for transitions from) |
List-II (Wavelengths (nm)) |
||
| A. | n2 = 3 to n1 = 2 | I. | 410.2 |
| B. | n2 = 4 to n1 = 2 | II. | 434.1 |
| C. | n2 = 5 to n1 = 2 | III. | 656.3 |
| D. | n2 = 6 to n1 = 2 | IV. | 486.1 |
Choose the correct answer from the options given below:
