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(B) In the Thomson model,the positive charge is uniformly distributed in a sphere of radius $R$. For an electron at a distance $r > R$,the electric fi

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$-13.6$

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(B) In the Thomson model,the positive charge is uniformly distributed in a sphere of radius $R$. For an electron at a distance $r > R$,the electric field is the same as that of a point charge $e$ at the center. The potential energy is $U(r) = -e^2 / (4 \pi \epsilon_0 r)$.Applying the Bohr quantization condition $mvr = n\hbar$ for $n=1$,we have $mvr = \hbar$. The centripetal force is provided by the electrostatic force: $mv^2 / r = e^2 / (4 \pi \epsilon_0 r^2)$,which gives $mv^2 = e^2 / (4 \pi \epsilon_0 r)$.Substituting $v = \hbar / (mr)$ into the force equation: $m(\hbar / mr)^2 = e^2 / (4 \pi \epsilon_0 r) \implies \hbar^2 / (mr^2) = e^2 / (4 \pi \epsilon_0 r) \implies r = 4 \pi \epsilon_0 \hbar^2 / (me^2) = a_0$,where $a_0$ is the Bohr radius $(0.53 \,\mathring A)$.Since the calculated radius $r = a_0 = 0.53 \,\mathring A$ is greater than the radius of the sphere $R = 0.25 \,\mathring A$,the electron orbits outside the sphere. In this region,the potential is equivalent to that of a point charge. Thus,the energy levels are the same as the standard Bohr model,and the ground state energy is $-13.6 \, eV$.

We consider the Thomson model of the hydrogen atom in which the proton charge is distributed uniformly over a spherical volume of radius $R = 0.25 \,\mathring A$. Applying the Bohr quantization condition in this model,the ground state energy (in $eV$) of the electron will be close to:

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