The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of `108.5mm`. The ground state energy of an electron of this ion will e
The wavelength of the third line of the Balmer series for a hydrogen atom is -
An ionic atom equivalent to hydrogen atom has wavelength equal to 1//4 of the wavelengths of hydrogen lines. The ion will be
The third line of the Balmer series spectrum of a hydrogen-like ion of atomic number Z equals to 108.5 nm . The binding energy of the electron in the ground state of these ions is E_(n) . Then
The series limit wavelength of the Balmer series for the hydrogen atom is
The first member of the Balmer series of hydrogen atom has wavelength of 6563 A. Calculate the wavelength and frequency of the second member of the same series.
The total energy of an electron in the ground state of hydrogen atom is - 13.6 eV . The potiential energy of an electron in the ground state of Li ^(2+) ion will be
The wavelength of the first line of the Balmer series of hydrogen atom is lamda . The wavelength of the corresponding line line of doubly ionized lithium atom is
The energy corresponding to second line of Balmer series for hydrogen atom will be :-
If the wavelength of the first line of the Balmer series of hydrogen atom is 656.1 nm the wavelngth of the second line of this series would be
