The electron in hydrogen atom is in the first Bohr orbit `(n = 1)`. The ratio of transition energies, `E(n-1 rarr n =3)` to `E(n=1 rarr n=2)`, is-

In hydrogen atom, the radius of n^(th) Bohr orbit is V_(n) . The graph Beetween log .(r_(n)/r_(1)) will be

Consider the follwoing two elerctronic transition possibilites in a hydrogen atom as pictured below : (1) The electron drops from third Bohr's orbit to second Bohr's obit followed with the next transition from second to first Bohr's orbit . (2) The electron drops from third Bohr's orbit to first Bohr's orbit directly . Show that : (a) The sum of the enrgies for the transitions n=3 to n=2 and n=2 to n=1 is equal to the energy of transiton for n=3 to n=1 . (b) Are wavelengths and frequencies of the emitted spectrum also additive in the same way as their energies are ?

The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth orbits from higher energy orbits respectively (as shown in figure) Maximum number of lines produced when an electron jumps from nth level to ground level is equal to (n(n-1))/(2) . For example, in the case of n = 4, number of lines produced is 6. (4 rarr 3, 4 rarr 2, 4 rarr 1, 3 rarr 2, 3 rarr 1, 2 rarr 1) . When an electron returns from n_(2) to n_(1) state, the number of lines in the spectrum will be equal to ((n_(2) - n_(1))(n_(2)-n_(1) +1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(2) then the difference may be expressed in terms of energy of photon as E_(2) - E_(1) = Delta E, lambda = (h c)/(Delta E) . Since h and c are constant, Delta E corresponds to definite energy, thus each transition from one energy level to another will prouce a higher of definite wavelength. THis is actually observed as a line in the spectrum of hydrogen atom. Wave number of the line is given by the formula bar(v) = RZ^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) Where R is a Rydberg constant (R = 1.1 xx 10^(7)) (i) First line of a series : it is called .line of logest wavelength. or .line of shortest energy.. (ii) Series limit of last of a series : It is the line of shortest wavelength or line of highest energy. The difference in the wavelength of the 2^(nd) line of Lyman series and last line of Bracket series in a hydrogen sample is

As the electron in Bohr orbit of hydrogen atom passesfrom state n = 2 to n = 1 , the kinetic energy K and potential energy U changes as

Energy of the electron in nth orbit of hydrogen atom is given by E_(n) =-(13.6)/(n^(2))eV . The amount of energy needed to transfer electron from first orbit to third orbit is

Energy of the electron in nth orbit of hydrogen atom is given by E_(n) =-(13.6)/(n^(2))eV . The amount of energy needed to transfer electron from first orbit to third orbit is

Calculate for a hydrogen atom the ionization potential, the first excitation potential and the wavelength of the resonance line (n'=2 rarr n=1) .