The kinetic energy of an electron in a Bohr orbit of the hydrogen atom is 8.57 × 10^-20 J. What is the radius of the orbit?

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The kinetic energy of an electron in a Bohr orbit is given by:$$K = \frac{-2\pi^2k^2\hbar^2}{2m}$$where:* K is the kinetic energy* k is the principal quantum number* ħ is the reduced Planck constant* m is the mass of the electronFor the hydrogen atom, the principal quantum number is 1, so we have:$$K = \frac{-2\pi^2(1)^2\hbar^2}{2m}$$Solving for r, we get:$$r = \frac{4\pi^2\hbar^2}{2mK}$$Plugging in the given values, we get:$$r = \frac{4\pi^2(1.0545718 imes10^{-34} ext{ J s})^2}{2(9.1093837 imes10^{-31} ext{ kg})(8.57 imes10^{-20} ext{ J})}$$$$r = 1.35 imes10^{-9} ext{ m}$$Therefore, the radius of the orbit is 1.35 × 10^-9 m.

The Kinetic Energy of an Electron in a Bohr Orbit