The Bohr model for the H-atom relies on

Question:

The Bohr model for the H-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge + q1, –q2 is modified to

$|F|=\frac{q_{1} q_{2}}{\left(4 \pi \epsilon_{0}\right)} \frac{1}{r^{2}}, r \geq R_{0}$

$=\frac{q_{1} q_{2}}{\left(4 \pi \epsilon_{0}\right)} \frac{1}{R_{0}^{2}}\left(\frac{R_{0}}{r}\right)^{\epsilon}, r \leq R_{0} \quad$ Calculate in such a case, the ground state energy of an $\mathrm{H}$-atom,

if ε = 0.1, R0 = 1Å.

Solution:

Case 1: When ε = 0.1, R0 = 1Å

R1 = 8 × 10-11

M = 0.08 nm

Velocity at ground level is v1 = 1.44 × 106 m/s

Case 2:

When n = 1 and v1 = h/mr1

Kinetic energy = 5.9 eV

Potential energy = -17.3 eV

Total energy = -11.4 eV

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