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2.18 In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 \dot{A}:

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 \dot{A} separation?

Answers (1)

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When potential energy  is zero at d' 1.06 \dot{A}  away,

The potential energy of the system =potential energy at d' -potential energy at d

PE=\frac{ke*p}{d_1}-27.2= \frac{9*10^{9}*(1.6*10^{-19})^2}{1.06*10^{-10}}=-13.6eV

Hence Potential energy, in this case, would be -13.6eV

Posted by

Pankaj Sanodiya

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