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Assuming the expression for radius of the orbit, derive an expression for total energy of an electron in hydrogen atom. |
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Answer» Solution :As per Rutherford.s model : `(mv^(2))/(r)=(1)/(4piepsi_(0)).(Ze^(2))/(r^(2))` `impliesmv^(2)=(1)/(4piepsi_(0)).(Ze^(2))/(r)`  TOTAL energy =P.E +K.E. `=-(1)/(4piepsi_(0)).(Ze^(2))/(r)+(1)/(2)mv^(2)=(-1)/(4piepsi_(0)).(Ze^(2))/(r)+(1)/(8piepsi).(Ze^(2))/(r)` `=-(1)/(2).(1)/(4piepsi_(0)).(Ze^(2))/(r)=-(1)/(8piepsi_(0)).(Ze^(2))/(r)` Negative sign implies that ELECTRON - NUCLEUS form a BOUND system. |
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