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Assuming the expressionfor theradiusof electronorbit ,the expression for thetotal energy of the electronin the stationaryorbit of hydrogen atom. |
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Answer» Solution :Consideran electronof MASS m and charge -e revolving round the nucleusof an atomof atomic number Z in the `n^(th)` orbit of radius .r.. Let v be the velocityof the electron.The electronpossessespotential energy , because , it is in the electrostatic fieldof the nucleus . The electron alsopossesses kinetic energy by VIRTUE of its motion. Potentialenergyof the electronis given by, `E_p`=(POTENTIALAT a distance r from the nucleus )(-e) `=1/(4piepsilon_0)[(ZE)/r](-e)` `E_p=-(Ze^2)/(4piepsilon_0r)`...(1) kinetic energy of the electronis given by , `E_k=1/2 mv^2`...(2) From Bohr.s postulate, `(mv^2)/r=1/(4piepsilon_0)[(Ze^2)/r^2]` `therefore mv^2=(Ze^2)/(4piepsilon_0r)` Substituting this value of `mv^2` in equation (2) `E_k=1/2 ((Ze^2)/(4piepsilon_0r))`...(3) Total ENERGYOF the electronrevolvingin 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]` `therefore E_n=-(Ze^2)/(8piepsilon_0r)` |
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