A simplified model of a hydrogen atom is that the electron cloud is a sphere of

Question

A simplified model of a hydrogen atom is that the electron cloud is a sphere of radius R with uniform charge density and total charge -e. (The actual charge density in the ground state is nonuniform.) See figure below a For the uniform density model, calculate the pola zability of atomic hydrogen n terms of R. Consider the case where the magnitude E of the applied electric field is much s aller than the electric e d e ure to ionize the atom. Suggestions for your analysis: Imagine that the hydrogen atom is inside a capacitor whose uniform field polarizes but does not accelerate the atom. Consider forces on the proton in the equilibrium situation, where the proton is displaced a distance s from the center of the electron cloud (s R in the diagram). (Use the following as necessary: R and o) b) For a hydrogen atom, R can be taken as roughly 1 x 10-20 m the Bohr model of the H atom gives R-0.5 x 10-10 m calculate a numerical value for the polarizability of atomic hydrogen. For comparison, the measured polarizability of a hydrogen atom is = 7.4 × 10-41 c . m/(N/C); see the note below. (Assume the hydrogen atom has a radius of 1 x 10-10 m.) C m/(N/C). (c) If the magnitude E of the applied electric field is 5.1 × 106 N/C, use the measured value of to calculate the shift s shown in the figure above. rm (d) For some purposes it is useful to model an atom as though the nucleus and electron cloud were connected by a spring. Use the measured value of to calculate the effective spring stiffness ks for atomic hydrogen. For comparison, measurements of Young's modulus show that the effective spring stiffness of the interatomic force in solid aluminum is about 16 N/m N/m (e) If were slx times as large, what would ks be? N/m

Explanation / Answer

Given,

radius = 8 cm ; q = 6 x 10^-8 C ;

B = 2 cm from surface

The correct statements are:

All the charges on the surface of the sphere will contribute to Eball at A

At location A, E ball is 0 N/C

Reason: since the charge spread uniformly over the surface of the conductor.

Electric field is ZERO inside a conductor

We know that the field is given by:

E = kq/r^2

r = 8 + 2 = 10 cm = 0.1 m

E = 9 x 10^9 x 6 x 10^-8/0.1^2 = 54000 N/C

Hence, E = 54000 N/C

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