Question

The ratio of Total energy and Potential energy of electron in Bohr orbit of $L{i}^{++}$ ion is:

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $4:1$

Answer 1

The correct option is B $1:2$

$K.E$ of Bohr $e^{-}=\frac{K\left(Ze\right)\left(e\right)}{2r}$
$P.E$ of Bohr $e^{-}=-\frac{K\left(Ze\right)\left(e\right)}{r}$
$T.E$ of Bohr $e^{-}=P.E+K.E=-\frac{K\left(Ze\right)\left(e\right)}{2r}$

$T.E:P.E=1:2$