Question

What is the kinetic energy of the electron in a Bohr hydrogen atom in the second excited state?

Answer 1

KE of the Bohr electron

m_ev²/2 ,

using columbs law and Newton's second law, the centrifugal force = electrostatic force

m_ev²/r = ke²/r²

Kinetic energy K = ke²/2r

r - Bohr radius is qunatised.

$r_n = \frac{n^2\not{h^2}}{m_eke^2}$ , where n is quantum number 1,2,3, . . . .

the smallest Bohr oribit $r_0 = \frac{\not{h^2}}{m_eke^2} = 0.0529 nm$

for the second excited state n= 3, n=1 is the ground stated

r₃ = 9*r_o = 4.761e-10 m

KE = 9.0e-7 *(1.6e-19)² / 2* 4.761e-10

       = 2.42 e-35 J