The energy of electron in the nth orbit of hydrogen atom is expressed as `E_(n) = (-13.6)/(n^(2))eV`. The shortest and longest wavelength of Lyman series will beA. `910 Å, 1213 Å`B. `5463 Å, 7858 Å`C. `1315 Å, 1530 Å`D. None of these

The energy of electron in the nth orbit of hydrogen atom is expressed

1.	The energy of electron in the nth orbit of hydrogen atom is expressed as `E_(n) = (-13.6)/(n^(2))eV`. The shortest and longest wavelength of Lyman series will beA. `910 Å, 1213 Å`B. `5463 Å, 7858 Å`C. `1315 Å, 1530 Å`D. None of these
Answer» Correct Answer - A `(1)/(lambda_(max)) = R[(1)/((1)^(2)) - (1)/((2)^(2))] rArr lambda_(max) = (4)/(3R) ~~ 1213 Å` and `(1)/(lambda_(min)) = R [(1)/((1)^(2)) - (1)/(oo)] rArr lambda_(min) = (1)/(R ) ~~ 910 Å`

Discussion

No Comment Found

Related InterviewSolutions

In hydrogen atom, when electron jupms from second to first orbit, then enrgy emitted isA. `-13.6 eV`B. `-27.2 eV`C. `-6.8 eV`D. None of these
In hydrogen atom, when electron jupms from second to first orbit, then enrgy emitted isA. `-13.6 eV`B. `-27. 2eV`C. `-6.8 eV`D. None of these
The enrgy of an atom (or ion) in the ground state is `- 54.4 eV ` .If may beA. `He^(+)`B. `Li^(2+)`C. hydrogenD. deuterium
Whenever a hydrogen atom emits a photon in the Balmer series .A. It may emit another photon in the Balmer seriesB. It must emit another photon in the lyman seriesC. the second photon, if emitted, will have a wavelength of about `122 nm`D. It may emit a second photon, the wavelength of this photon cannot be predicted
Electrons of energy `12.09 eV` can excite hydrogen atoms . To which orbit is the electron in the hydrogen atom raisd and what are the wavelengths of the radiations emitted as it dropes back to the ground state?
Find the quantum number `n` corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelengths state that ion emits two photons in succession with wavelengths`1026.7 and 304Å. (R = 1.096 xx 10^(7)m^(_1)`
Find the quantum number `n` corresponding to the excited state of `He^(+)` ion, if on transition to the ground state that ion emits two photons in succession with wave lengths `108.5` and `30.4 nm`.
Electron of energies 10.20 eV and 12.09 eV` can cause radiation to be emitted from hydrogen atoms . Calculate in each case, the principal quantum number of the orbit to which electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.
A hydrogen atom ia in excited state of principal quantum number `n` . It emits a photon of wavelength `lambda` when it returnesto the ground state. The value of `n` isA. `sqrt(lambda R (lambda R - 1))`B. `sqrt((lambda (R - 1))/(lambda R))`C. `sqrt((lambda R)/(lambda R - 1))`D. `sqrt lambda (R - 1)`
The value of wavelength radiation emitted by an `H`-atom, if atom is excited to state with principal quantum number four is :A. `(5R)/(36)`B. `(16)/(15R)`C. `(36)/(5R`D. `(3R)/(16)`

The energy of electron in the nth orbit of hydrogen atom is expressed as `E_(n) = (-13.6)/(n^(2))eV`. The shortest and longest wavelength of Lyman series will beA. `910 Å, 1213 Å`B. `5463 Å, 7858 Å`C. `1315 Å, 1530 Å`D. None of these

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment