Centripetal acceleration?

Question

For a hydrogen atom in the n = 3 Bohr orbital, what is the

centripetal acceleration of the electron?

Answer 1

The energy of an electron in the first orbital is -2.18 x 10^-18 J.

En = E1/n^2 so

E3 = E1/3^2 or -2.18 x 10^-18 J/9 = -2.42 x 10^-19 J.

rn = r1 x n^2 so

r3 = r1 x 3^2 = 5.29E-11 x 9 = 4.76E-10 m

Kinetic energy = mv^2/2, so v = root of 2E/m

v = root (2)(-2.42E-19)/(9.11E-31) = 7.29E5 m/s

ac = v^2/r = (7.29E5 m/s)^2/4.76E-10 m = 1.12E21 m/^2