an electron in a hydrogen LIKE species is associated with a wave length of 6.67 A⁰. wave no. of lyman's series of that specie is 438708 cm^-1. the species is excited electrically to some higher quantum state which has 3.808 eV more than the 2nd excited state. R = 109677 cm^-1

 

1 what is Kinetic energy of electron? 3.38 eV
2 what is PE of e- ? -6.78 eV
3 what will b wavelength of photon emitted during the shortest wavelength of transition from the mentioned excited state? 237.4 A⁰

1) To find the kinetic energy, use the formula =h/mv and then calculate kinetic energy by 1/2mv².

 = 6.67 x 10^-10 = 6.63 x 10^-34

                                    ----------------------------------------

                            9.1 x 10^-31      x    v

 

which on simplification gives v=0.109 x 10⁷  m/s.

K.E. = 1/2mv²   where m = mass of electron

On putting the values and getting the answer in joule as 

5.4 x 10^-19    J , convert it into eV to get 3.38 eV.

 

2) Now,

    as K.E. = kZe²

                   ------

                    2r  

and P.E.= - kZe²  / r

we have P.E.= - 2 K.E.

 

Thus, P.E.= -6.78 eV.

 

3)       Wavenumber

      --------------------------    =   Z² {1- 1/ ²}

         Rydberg's constant

From this u will get Z=2.

 

Now, E= -13.6 Z²/N²  EV

 

Using this or rydberg's equation, wavelength of photon emitted during the shortest wavelength of transition from the excited state can be found out easily.

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