Find the value of wave number `(overset-v)` in terms of Rydberg's constant, when transition of electron takes place between two lvels of `He^(+)` ion whose sum is 4 and difference is 2.

Calculate the wavelength emitted during the transition of an electron in between two level of Li^(2+) ion whose sum is 4 and difference is 2

The only element in the hydrogen atom resides under ordinary condition on the first orbit .When energy is supplied the element move to hjgher energy ornbit depending on the lower of energy absioerbed .When this electron to may of the electron return to any of the lower orbits, it emit energy Lyman series is formed when the electron to the lowest orbit white Balmer series ids formed when the electron returns to the second orbit similar Paschen Brackett, and Pfund series are formed when electron return to the third fourth , and fifth arbit from highest energy orbits, respectively Maximum number of liner produced is equal when as electron jumps from nth level to ground level is equal to (n(n - 1))/(2) If teh electron comes back from the energy level having energy E_(2) to the energy level having energy E_(1) then the difference may be expresent in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = hc//Delta E Since h and c are constants Delta E coresponding to definite energy , thus , each transition from one energy level to unother will produce a light of definite wavelem=ngth .This isd actually observed as a line in the spectrum of hydrogen atom Wave number of line is given by the formula bar v = RZ^(2)((1)/(n_(1)^(2))- (1)/(n_(12)^(2))) Where R is a Rydherg constant The wave number of electromagnetic radiation emitted during the transition of elecvtron in between the two levels of Li^(2+) ion whose pricipal quantum numbner sum is 4 and difference is 2 is

The wave number of electromagnetic radiations emitted during the transition of electron in between two levels of Li^(2+) ion having sum of the principal quantum numbers 4 and difference is 2 ,will be: ( R_(H) = Rydberg constant )

The angular momentum of an electron in a Bohr's orbit of He^+ is 3.1652 xx 10^-34 kg-m^2//sec . What is the wave number in terms of Rydberg constant (R ) of the spectral line emitted when an electron falls this level to the first excited state. ["Use" h = 6.626 xx 10^-34 Js] .

The angular momentum of an electron in a Bohr's orbit of He^(+) is 3.1652xx10^(-34) kg- m^(2) /sec. What is the wave number in terms of Rydberg constant (R ) of the sepectral line emitted when an electron falls from this level to the first excited state.l [ Use h =6.626xx10^9-34) Js]

For a single electron atom or ion the wave number of radiation emitted during the transition of electron from a higher energy state (n = n_(2)) to a lower energy state (n=n_(1)) is given by the expression: bar(v) = (1)/(lambda) = R_(H).z^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) ...(1) where R_(H) = 2(pi^(2)mk^(2)e^(4))/(h^(3)c) = Rydberg constant for H-atom Where the terms have their usual meanings. Considering the nuclear motion, the most accurate expression would have been to replace mass of electron (m) by the reduced mass (mu) in the above expression, defined as mu = (m'.m)/(m'+m) where m'= mass of nucleus For Lyman series: n_(1) =1 (fixed for all the lines) while n_(2) = 2,3,4 ... for successive lines i.e. 1^(st), 2^(nd),3^(rd) ... lines, respectively. For Balmer series: n_(1) = 2 (fixed for all the lines) while n_(2) = 3,4,5 ... for successive lines. If proton in H-nucleus be replaced by positron having the same mass as that of electron but same charge as that of proton, then considering the nuclear motion, the wavenumber of the lowest energy transition of He^(+) ion in Lyman series will be equal to