When the electron in a hydrogen atom jumps from the second orbit to the first orbit , the wavelength of the radiation emitted is lamda . When the electron jumps from the third orbit to the first orbit , of the same atom , the wavelength of the emitted radiation would be
When a hydrogen atom is excited from ground state to first excited state then
When a hydrogen atom is excited from ground state to first excited state, then
The wave number of the first line in the Balmer series of hydrogen atom is 15200 cm^(-1) . What is the wave number of the first line in the Balmer series of Be^(3+) ?
Determine the number of enegry states in the hydrogen atom corresponding to the principle quantum number n = 2 and calculate the energies of these states. Strategy: An energy state of the hydrogen atom can be respresented by the wave function psi_(n,l,m_(1)) , which can be specified by sunstituting the values of the three quantum numbers.
If the wave number of the first line in the Balmer series of hydrogen atom is 1500cm^(-1) , the wave number of the first line of the Balmer series of Li^(2+) is
Ratio of the wave number of the first line of Lyman series of hydrogen atom to the first line of Balmer series of He^(+) ion will be
For emission line of atomic hydrogen from n_(i)=8 to n_(f)=n, the plot of wave number (barv) against ((1)/(n^(2))) will be (The Rydberg constant, R_(H) is in wave number unit)
The wave number of the first line in the balmer series of Be^(3+) ?
