# 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 = $\frac{Kz{e}^{2}}{2r}=\frac{9×{10}^{9}×1×\left(1.6×{10}^{-19}{\right)}^{2}}{2×r}=\frac{11.52×{10}^{-29}}{r}$
9.7*10-19 =$\frac{11.52×{10}^{-29}}{r}$
r = 1.19*10-10 m

Regards

• 8
What are you looking for?