The wavelength (in Å) of an emission line obtained for `Li^(2+)` during an electronic transition `n_(2) = 2` and n = 1 is (R = Rydberg constant)

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :

The electronic transition from n=2 to n=1 will produce shortest wavelength in:

The wave number of the first emission line in the Balmer series of H - Spectrum is (n)/(36)R . The value of 'n' is. (R = Rydberg constant):

The wave number of the first emission line in the Balmer series of H - Spectrum is (n)/(36)R . The value of 'n' is. (R = Rydberg constant):

What will be the wavelength of the light emitted due to a transition of electron from n = 3 orbit to n = 2 orbit in hydrogen atom? Given: the Rydberg constant for hydrogen atom is R_(H) = 1.09 xx 10^(7) m^(-1) .

One of the lines in the emission spectrum of Li^(2+) has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 to n=x. Find the value of x.

One of the lines in the emission spectrum of Li^(2 +) has the same wavelength as that of the second line of Balmer series in hydrogen spectrum. The electronic transition correspnding to this line is.

One of the lines in the emission spectrum of Li^(2 +) has the same wavelength as that of the second line of Balmer series in hydrogen spectrum. The electronic transition correspnding to this line is.