The atomic spectrum of `Li^(+2)` ion arises due to the transition of an electron from `n_(2)` to `n_(1)` if `n_(1) +n_(2)=4` and `(n_(2)-n_(1))` =2 then the wavelength of `3^(rd)` line of this series in `Li^(+2)` ion will be

The atomic spectrum of Li^(2+) arises due to the transition of an electron from n_(2) to n_(1) level. If n_(1) +n_(2) is 4 and n_(2)-n_(1) is 2, calculate the wavelength (in nm) of the transition for this series in Li^(2+)

The transition of an electron from n_(2)=5,6 ….. To n_(1) =4 gives rise to

The atomic structure of He^(+) arises due to transition from n_(2) to n_(1) level. If n_(1)+n_(2) is 3 and n_(2)-n_(1) is 1. Find the lambda in nm of transition for this series in He^(+) in nm.

The vale of wavelength radiation emitted due to transition of electrons from n=4 to n=2 state in hydrogen atom will be

When electron jumps from higher orbit to lower orbit, then energy is radiated in the form of electromagnetic radiation and these radiations are used to record the emission spectrum Energy of electron may be calculated as E =- (2pi^(2)m_(e)Z^(2)e^(4))/(n^(2)h^(2)) Where, m_(e)= rest mass of electron DeltaE = (E_(n_(2))-E_(n_(1))) = 13.6 xx Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]eV per atom This equation was also used by Rydberg to calculate the wave number of a particular line in the spectrum bar(v) = (1)/(lambda) = R_(H)Z^(2) [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]m^(-1) Where R_(H) = 1.1 xx 10^(7)m^(-1) (Rydberg constant) For Lyman, Balmer, Paschen, Brackett and Pfund series the value of n_(1) = 1,2,3,4,5 respectively and n_(2) =oo for series limit. If an electron jumps from higher orbit n to ground state, then number of spectral line will be .^(n)C_(2) . Ritz modified the Rydberg equation by replacing the rest mass of electron with reduced mass (mu) . (1)/(mu) = (1)/(m_(N))+ (1)/(m_(e)) Here, m_(N)= mass of nucleus m_(e)= mass of electron Answer the following questions The emission spectrum of He^(+) involves transition of electron from n_(2) rarr n_(1) such that n_(2)+n_(1) = 8 and n_(2) -n_(1) = 4 . what whill be the total number of lines in the spectrum?

In the atomic spectrum of hydrogen, the spectral lines pertaining to electronic transition of n = 4 to n = 2 refers to

An electron in C^(5+) ion during the transition from n = 3 to n = 1 emits light of wavelength

Taking the wavelength of first Balmer line in the hydrogen spectrum (n=3" to "n=2) as 660 nm, then the wavelength of 2^("nd") Balmer line in the same spectrum (n=4" to "n=2) will be