Arun
Last Activity: 6 Years ago
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
Regards
Arun