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

The series obtained by the transition of the electron from n_(2)=5,6,..." to "n_(1)=4 is called _______ series.

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 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) 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

In the spectrum of singly ionized helium , the wavelength of a line abserved is almost the same as the first line of Balmer series of hydrogen . It is due to transition of electron from n_(1) = 6 to n_(2) = ^(''**'') . What is the value of (''**'') .

In the spectrum of singly ionized helium , the wavelength of a line abserved is almost the same as the first line of Balmer series of hydrogen . It is due to transition of electron from n_(1) = 6 to n_(2) = ^(''**'') . What is the value of (''**'') .

The transition of electron from n = 3 to n = 1 in H atom produces

The transition of electron from n = 3 to n = 1 in H atom produces

Transition of electron from n=3 to n=1 level results in

The transition of the electron takes place from n = 2 orbit to n = 1 orbit. Which of the following gives the shortest wavelength ?