Question

There is an electron in the Bohr model in orbit 2. n=2. Now using de Brogile standing wave perspective so the electron in this orbit. Calculate the time required for the electron to complete one revolution.

Calculate the current in amperes corresponding to the electron motion in orbit 2.

What is the magnitude of the magnetic field produced at the nucleus due to electron circulation? Know there is a current now. There is a circular current. There will be a magnetic field.

What is the wavelength and linear momentum of the photon emitted with this electron makes a transition to the ground state.

Answer 1

v = 2.165*10⁶*(z/n) m/s = 1.08*10⁶ m/s

r = 0.53*10^-10(n²/Z) = 2.12*10^-10 m

Circumference of orbit = 2πr = 13.32*10^-10m

Time taken for one revolution (t) = circumference/velocity = 12.34*10^-16 s

Current = e/t = (1.602*10^-19)/(12.34*10^-16) = 6.24*10^-4 A

Magnetic field at nucleus = μ₀I/2r = 1.85 T

1/λ = 10973731[(1/1) - (1/4)]

=> λ = 1.22*10^-7 m

momentum (mv) = h/λ = 6.626*10^-34/1.22*10^-7 = 5.43*10^-27 kgm/s