Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is:

The de Broglie wavelength is given by λ = h/p, where h is Planck's constant and p is the momentum of the electron. In a hydrogen atom, the electron's momentum is related to its angular momentum, which is quantized: mvr = nħ, where m is the electron's mass, v is its velocity, r is the radius of its orbit, n is the principal quantum number, and ħ is the reduced Planck's constant (h/2π). The second excited state corresponds to n = 3. We can solve for v and substitute it into the de Broglie wavelength equation.

1. Find the velocity (v):
   mvr = nħ
   v = nħ / mr

2. Substitute into the de Broglie equation:
   λ = h/p = h/(mv) = h/((nħ/r) ) = hr/(nħ) = 2πr/n

3. Plug in the values:
   Given r = 4.65 Å and n = 3,
   λ = 2π(4.65 Å)/3 ≈ 9.7 Å

Therefore, the de Broglie wavelength of the electron in the second excited state is approximately 9.7 Å.