the kinetic energy of an e- in H like atom is 6.04eV. the area of the 3rd Bohr orbit to which this e- belong is??

Dear Student,

Kinetic energy = 6.04 eV = 9.7*10^-19 J
Kinetic energy of electron = $\frac{Kz{e}^2}{2r}$
K = constant = 9*10⁹ Nm²/C²
​z = 1 (for hydrogen like atom)
e = 1.6*10^-19 C
Kinetic energy of electron = $\frac{Kz{e}^2}{2r}=\frac{9\times {10}^9\times 1\times (1.6\times {10}^{-19}{)}^2}{2\times r}=\frac{11.52\times {10}^{-29}}{r}$
9.7*10^-19 =$\frac{11.52\times {10}^{-29}}{r}$
r = 1.19*10^-10 m
$area = 4{\mathrm{\pi r}}^2=4\times 3.14\times {\left(1.19\times {10}^{-10}\right)}^2=17.8\times {10}^{-20} {\mathrm{m}}^2$

Regards