In H-atom , if x is the radius of the first Bohr orbit, then what is the de Broglie wavelength of an electron in 3rd orbit?

For a hydrogen atom, radius of nth orbit, rn = n2h2/ 4π2mZe2 

   Therefore, r1/r3 = 12/32

   r3 = 9/r1

  = 9/x

Now, according to De Broglie; angular momentum of electron in 3rd orbit is:

mvr3 = 3h/2π   or  h/mv = 2πr3/3

And

λ = h/mv   where, λ is the de Broglie wavelength.

Therefore,

λ = 2πr3/3

  = 2π x 9/3x

  = 6πx