Advertisements
Advertisements
प्रश्न
Determine the series limit of Balmer, Paschen and Brackett series, given the limit for Lyman series is 911.6 Å.
Advertisements
उत्तर
Data: `lambda_("L"∞)` = 911.6 Å
For hydrogen spectrum, `1/lambda = "R"_"H"(1/"n"^2 - 1/"m"^2)`
∴ `1/lambda_("L"∞) = "R"_"H"(1/1^2 - 1/∞) = "R"_"H"` ...(1)
as n = 1 and m = ∞
`1/lambda_("B"∞) = "R"_"H"(1/4 - 1/∞) = "R"_"H"/4` .....(2)
as n = 2 and m = ∞
`1/lambda_("Pa"∞) = "R"_"H"(1/9 - 1/∞) = "R"_"H"/9` .....(3)
as n = 3 and m = ∞
`1/lambda_("Br"∞) = "R"_"H"(1/16 - 1/∞) = "R"_"H"/16` ......(4)
as n = 4 and m = ∞
From Eqs. (1) and (2), we get,
`lambda_("B"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//4) = 4`
∴ `lambda_("B"∞) = 4 lambda_("L"∞) = (4)(911.6)` = 3646 Å
This is the series limit of the Balmer series.
From Eqs. (1) and (3), we get,
`lambda_("Pa"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//9)` = 9
∴ `lambda_("Pa"∞) = 9lambda_("L"∞) = (9)(911.6)` = 8204 Å
This is the series limit of the Paschen series.
From Eqs. (1) and (4), we get,
`lambda_("Br"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//16)` = 16
∴ `lambda_("Br"∞) = 16 lambda_("L"∞) = (16)(911.6)` = 14585.6 Å
This is the series limit of the Brackett series.
APPEARS IN
संबंधित प्रश्न
In both β− and β+ decay processes, the mass number of a nucleus remains the same, whereas the atomic number Z increases by one in β− decay and decreases by one in β+ decay. Explain giving reason.
Determine the series limit of Balmer, Paschen, and Pfund series, given the limit for Lyman series is 912 Å.
In Balmer series, wavelength of first line is 'λ1' and in Brackett series wavelength of first line is 'λ2' then `lambda_1/lambda_2` is ______.
In the figure below, D and E respectively represent
Which of the following transition will have highest emission wavelength?
Which of the following is TRUE?
Let the series limit for Balmer series be 'λ1' and the longest wavelength for Brackett series be 'λ2'. Then λ1 and λ2 are related as ______.
Which of the following transition will have highest emission frequency?
In hydrogen atom, the product of the angular momentum and the linear momentum of the electron is proportional to (n = principal quantum number) ____________.
In hydrogen spectrum, the series of lines obtained in the ultraviolet region of the spectrum is ____________.
If the mass of the electron is reduced to half, the Rydberg constant ______.
Each element is associated with a ______.
Absorption line spectrum is obtained ______.
To produce an emission spectrum of hydrogen it needs to be ______.
The first four spectral lines in the Lyman series of a H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4 Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.
Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H and 2H. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass µ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here µ = meM/(me + M) where M is the nuclear mass and m e is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in 1H and 2H. (Mass of 1H nucleus is 1.6725 × 10–27 kg, Mass of 2H nucleus is 3.3374 × 10–27 kg, Mass of electron = 9.109 × 10–31 kg.)
The first three spectral lines of H-atom in the Balmer series are given λ1, λ2, λ3 considering the Bohr atomic model, the wavelengths of the first and third spectral lines `(lambda_1/lambda_3)` are related by a factor of approximately 'x' × 10–1. The value of x, to the nearest integer, is ______.
The frequencies for series limit of Balmer and Paschen series respectively are 'v1' and 'v3'. If frequency of first line of Balmer series is 'v2' then the relation between 'v1', 'v2' and 'v3' is ______.
In the hydrogen atoms, the transition from the state n = 6 to n = 1 results in ultraviolet radiation. Infrared radiation will be obtained in the transition.
Find the wavelength and wave number of the first member of the Balmer series in Hydrogen spectrum. (`R =1.097xx10^7m^(-1)`)
The de-Broglie wavelength of the electron in the hydrogen atom is proportional to ______.
