In a hydrogen atom, the electron and proton are bound at a distance of about $0.53\; \mathring A:$
$(a)$ Estimate the potential energy of the system in $eV$, taking the zero of the potential energy at infinite separation of the electron from proton.
$(b)$ What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)?$
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06\;\mathring A$ separation?
In a hydrogen atom, the electron and proton are bound at a distance of about $0.53\; \mathring A:$
$(a)$ Estimate the potential energy of the system in $eV$, taking the zero of the potential energy at infinite separation of the electron from proton.
$(b)$ What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)?$
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06\;\mathring A$ separation?
$(a)$ Potential at infinity is zero. Potential energy of the system, $p-e$ $=$ Potential energy at infinity - Potential energy at distance, $d =0-\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} d}$
Where, $\epsilon_{0}$ is the permittivity of free space
$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \,N\,m ^{2} \,C ^{-2}$
$\therefore$ Potential energy $=0-\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{0.53 \times 10^{10}}$$=-43.7 \times 10^{-19}\,J$
since $1.6 \times 10^{-19}\, J =1 \,eV$
$\therefore$ Potential energy $=-43.7 \times 10^{-19}=\frac{-43.7 \times 10^{-19}}{1.6 \times 10^{-19}}$$=-27.2\, eV$
Therefore, the potential energy of the system is $-27.2 \,eV.$
$(b)$ Kinetic energy is half of the magnitude of potential energy.
Kinetic energy $=\frac{1}{2} \times(-27.2)=13.6 \,eV$
Total energy $=13.6-27.2=13.6 \,eV$
Therefore, the minimum work required to free the electron is $13.6 \,eV.$
$(c)$ When zero of potential energy is taken, $d_{1}=1.06\, \mathring A$
$\therefore$ Potential energy of the system $=$ Potential energy at $d _{1}-$ Potential energy at $d$ $=\frac{q_{1} q_{2}}{4 \pi \epsilon_{0} d_{1}}-27.2 \,eV$
$=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{1.06 \times 10^{-10}}-27.2\, eV$
$=21.73 \times 10^{-19}\, J -27.2 \,eV$
$=13.58\, eV -27.2\, eV$
$=-13.6 \,eV$
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