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Q.
What is the K.E and P.E of electron in the first orbit of hydrogen atom? Given $e =1.6 \times 10^{-19} C , r =0.53 \times 10^{-10}$
$m$
Atoms
Solution:
K.E. of electron $=\frac{ KZe ^{2}}{2 r }=\frac{ Ke ^{2}}{2 r } (\because Z =1)$
$K.E =\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{2 \times 0.53 \times 10^{-10}}$ Joule
$=\frac{9 \times 2.56 \times 10^{-19}}{1.06 \times 1.6 \times 10^{-19}} eV =13.58 eV$
P.E of electron $=\frac{- K Ze ^{2}}{ r }=-2( K , E )$
$=2 \times(13.58)=-27.16 eV$