and the term in the parenthesis is positive and energy is absorbed. On the other hand in case of emission spectrum n i > n f , ∆ E is negative and energy is released. The expression ( . ) is similar to that used by Rydberg ( .
) derived empirically using the experimental data available at that time. Further, each spectral line, whether in absorption or emission spectrum, can be associated to the particular transition in hydrogen atom. In case of large number of hydrogen atoms, different possible transitions can be observed and thus leading to large number of spectral lines. The brightness or intensity of spectral lines depends upon the number of photons of same wavelength or frequency absorbed or emitted.
Problem . What are the frequency and wavelength of a photon emitted during a transition from n = state to the n = state in the hydrogen atom? Since n i = and n f = , this transition gives rise to a spectral line in the visible region of the Balmer series. From equation ( .
) E = = . . J J It is an emission energy The frequency of the photon (taking energy in terms of magnitude) is given by = . × Hz Problem .
Calculate the energy associated with the first orbit of He + . What is the radius of this orbit? E Z J ( . ) atom – For He + , n = , Z = E ( .
)( ) . J J The radius of the orbit is given by equation ( . ) r nm Z = ( . ) Since n = , and Z = r nm nm n = = ( .
) . 02645 . . Limitations of Bohr’s Model Bohr’s model of the hydrogen atom was no doubt an improvement over Rutherford’s nuclear model, as it could account