Saved Bookmarks
| 1. |
Derive an experession for the total energy of an electron in stationary state of hydrogen atom. Assuming the expression for the radius. |
|
Answer» Solution :Consider an electron of mass m and charge -e REVOLVING around the nucleus of an atom of atomic number Z in the `n^(th)` orbit of radius r. Let v be the velocity of the electron. The electron possesses potential energy, because, it is in the electrostatic field of the nucleus. The electron also possesses kinetic energy by virtue of its motion. Potential energy of the electron is given by, `E_p` = (potential at a distance r from the nucleus)(-e) `=1/(4piepsilon_0) ["Ze"/r](-e)` `E_P=-(Ze^2)/(4piepsilon_0r)` ...(1) Kinetic energy of the electron is given by, `E_k=1/2 mv^2` ...(2) From Bohr.s postulate , `(mv^2)/r=1/(4piepsilon_0) [(Ze^2)/r^2] ` `thereforemv^2 = (Ze^2)/(4piepsilon_0r)` Substituting this value of `mv^2`in equation(2) `E_k=1/2 ((Ze^2)/(4piepsilon_0 r))` ...(3) Total energy of the electronrevolving in the `n^(th)`orbit is given by ` E_n=E_p + E_k` `E_n=-(Ze^2)/(4piepsilon_0r) +1/2 ((Ze^2)/(4piepsilon_0r))` Using(1) and (2) `=(Ze^2)/(4piepsilon_0r) [(-1)/1 + 1/2]` `=(Ze^2)/(4piepsilon_0r) [-1/2]` `thereforeE_n=-(Ze^2)/(8piepsilon_0r)` . The radius of `n^(th)`permitted orbit of the electronis given by `r=(epsilon_0 n^2 h^2)/(pi m Ze^2) ` Substitutingthis value ofr in the equation , `E_n = -(Ze^2)/(8 pi epsilon_0 r)` we get ,`E_n =-(Ze^2)/(8pi epsilon_0) xx(pimZe^2)/(epsilon_0 n^2 h^2)` i.e.,`E_n=(-Z^2 me^4)/(8epsilon_0^2n^2h^2)`for hydrogenlike atoms. For HYDROGEN atom , put Z=1 `therefore` Total energy of the ELECTRONIN the `n^(th)`orbit of hydrogen atom is `E_n=(-me^4)/(8epsilon_0^2 n^2 h^2)` NOTE : The Total energy of the electron in the firstorbit is `E_1 = (-me^4)/(8epsilon_0^2 h^2)`=-13.6 eV |
|