Question

A Hydrogen atom has one proton in the nucleus and one electron in the shell. In a classic model of the atom, in a certain state, this electron is in a circular orbit around the nucleus with an angular momentum of 9.495e-34 Js.
What is the radius of the orbit?
4.30×10^-9m   

What is the speed of the electron at this radius?

What is the kinetic energy of the electron at this radius?

What is the kinetic energy in electron-volts?

Answer 1

(b)

First we will find out speed of electron.

Angular momentum, L = 9.495*10^(-34) Js

Magnetic force = Centripetal force on atom,

ke^2 / R^2 = mv^2 / R

ke^2 = (mvR)*R

and mvR = L

9*10^9*(1.6*10^(-19))^2 = 9.495*10^(-34)*v

v = 2.42*10^5 m/s

(a)

mvR = 9.495*10^(-34)

R = 9.495*10^(-34) / 9.1*10^(-31)*2.42*10^5

R = 4.29*10^(-10) m

(c)

kinetic energy of the electron

KE = (1/2)mv^2

KE = (1/2)*9.1*10^(-31)*(2.42*10^5)^2

KE = 2.67*10^(-20) J

(d)

kinetic energy in electron-volts,

KE = 2.67*10^(-20) / 1.6*10^(-19)

KE = 0.1674 eV