Advertisements
Advertisements
प्रश्न
Determine the series limit of Balmer, Paschen, and Pfund series, given the limit for Lyman series is 912 Å.
Advertisements
उत्तर
Data: `lambda_("L"∞)` = 912 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_("Pf"∞) = "R"_"H"(1/25 - 1/∞) = "R"_"H"/25` ......(4)
as n = 5 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)(912)` = 3648 Å
This is the series limit of the Balmer series.
rom Eqs. (1) and (3), we get,
`lambda_("Pa"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//9)` = 9
∴ `lambda_("Pa"∞) = 9lambda_("L"∞) = (9)(912)` = 8208 Å
This is the series limit of the Paschen series.
From Eqs. (1) and (4), we get,
`lambda_("Pf"∞)/lambda_("L"∞) = "R"_"H"/("R"_"H"//25)` = 25
∴ `lambda_("Pf"∞) = 25 lambda_("L"∞) = (25)(912)` = 22800 Å
APPEARS IN
संबंधित प्रश्न
What is the shortest wavelength present in the Paschen series of spectral lines?
An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom? To which series these lines correspond?
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.
The energy (in eV) required to excite an electron from n = 2 to n = 4 state in hydrogen atom is ____________.
In hydrogen spectrum, which of the following spectral series lies in ultraviolet region?
The radii of the first four Bohr orbits of hydrogen atom are related as ____________.
In hydrogen spectrum, the series of lines obtained in the ultraviolet region of the spectrum is ____________.
In hydrogen spectrum, the wavelengths of light emitted in a series of spectral lines is given by the equation, `1/lambda` = R `(1/4^2 - 1/"n"^2)`, where n = 5, 6, 7...... and 'R' is Rydberg's constant. Identify the series and wavelength region.
Continuous spectrum is produced by ______.
An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. What is the wave number of the emitted radiations? (R = Rydberg's constant)
Absorption line spectrum is obtained ______.
What is the minimum energy that must be given to a H atom in ground state so that it can emit an Hγ line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such Hγ photon?
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.
Determine the shortest wavelengths of Balmer and Pasch en series. Given the limit for the Lyman series is 912 Å.
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 ______.
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 ______.
Calculate the wavelength of the first two lines in the Balmer series of hydrogen atoms.
What condition of an atomic gas or vapour produces an emission line spectrum?
What type of spectrum is produced when white light passes through an atomic gas or vapour and dark lines appear?
What is passed through an atomic gas or vapour to produce an absorption spectrum?
What does an atomic gas or vapour absorb when white light passes through it?
