An old model of a hydrogen atom has the charge of the proton uniformly distributed over a sphere of radius , with the electron of charge -e and mass at its center.
The force exerted on the electron if it is moved by a distance of is equal to its charge times the local electric field.
F = -eE (1)
Where, E is electric field and e is charge.
Here we have, radius of sphere is
Mass of sphere is m.
By Gauss law we have,
Where, q is enclosed charge and is permittivity.
We need to determine the electric field at r distance from the nucleus, therefore the enclosed charge is calculated as the product of the volumes of spheres with r and a multiplied by the proton charge, as follows:
So, the electric field at distance of r is given by:
Now, by substituting the value of in equation (1) we get,
Hence, the force on the electron is .
Now, from the Newton’s second law of motion we have,
Now, by step 3 we have,
So, we get
Hence, the angular frequency is given by .
A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of 0.85 eV makes a transition to a state with an excitation energy (the difference between the energy of the state and that of the ground state) of . (a) What is the energy of the photon emitted as a result of the transition? What are the (b) higher quantum number and (c) lower quantum number of the transition producing this emission?
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