Question

assume that electron in he ion is excited to the second orbital(n=2) calculate the following: the radius of the orbit,the velocity of the electron,the potential energy of the electron, the kinetic energy of the electron

Answer 1

Bohr's radius for the nth Orbit:  (n = principal quantum number)

Let
   R = Bohr' radius for an atom of atomic number Z,
   n = orbit number = principal quantum number  =   2
   h = Planck's constant = 6.626 * 10⁻³⁴ units
   K = 1/(4πε₀) = 9 * 10⁹ N-m²/C²  = Coulomb's constant
   Z = 2  for Helium
   m = mass of an electron = 9.1 * 10⁻³¹ kg
   e = charge on the electron = 1.602 * 10⁻¹⁹ C

1) centripetal force = electrostatic attraction between an electron and protons.
         m v² / R  =  K (Z*e) * e / R²
      => v² = K Z e² / (m R)      --- (1)

2) Angular momentum = m v R = n h / 2π          (integral multiple of  h/2π)
         =>  v = n h / (2 π m R)    --- (2)
   
3)  from (1) and (2):
        n² h² / (4π² m² R²) = K Z e² / (m R)
    => R = n² h² / (4π² m K e² Z)      --- (3)

4)  So speed of electron (linear along the circular orbit)  by substituting value of R,
    =>  v = (2 π  K e² Z) / (n h)

5)  Potential energy of the electron:
         We ignore gravitational potential energy here.
       PE = - K * Z * e * e / R  = - K Z e² / R  --- (4)
               = - [4 π² m K² Z² e⁴ ] / (n² h²)

6) Kinetic energy of electron:
         => 1/2 * m * v² = (π m * R e² Z ) / (n h)
               = [ 2 π² K² Z² e⁴ m ] / (n² h²)  =  - P.E / 2
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7)  The total energy of the electron :  (a simple formula)
         KE + PE =  P.E / 2
         Total energy = - 13.6 Z² / n²  eV = - 13.6 e V

So K.E. = 13.6 e V        and  P.E. = - 27.2 e V

Bohr's Radius of Hydrogen atom R for n = 1 is  0.529 °A
     So for Helium in n =2,    R =  n² * 0.529 / Z  °A  = 1.058  °A

Speed of electron in Hydrogen (n = 1) is  v = 2,185 km/s      (≈ speed of light / 137)
       so for Helium in n = 2,  v =  z * 2,185 / n   km/s
                   So  v =  2, 185  km/s