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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.
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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`
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संबंधित प्रश्न
The ground state energy of hydrogen atom is – 13∙6 eV. If an electron makes a transition from an energy level – 1∙51 eV to – 3∙4 eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.
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[Given R = 1.1 ✕ 107 m−1]
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The transitions A, B and C respectively represents
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