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In a hydrogen atom, the electron and proton are bound at a distance of about $0.53 Å$.
(a) Estimate the potential energy of the system in $\mathrm{eV}$, 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 $1.06 Å$ separation?
(a) Estimate the potential energy of the system in $\mathrm{eV}$, 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 $1.06 Å$ separation?
Solution:
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Verified Answer
(a) Given, $\mathrm{q}_1=+1.6 \times 10^{-19} \mathrm{C}$,
$\mathrm{q}_2=-1.6 \times 10^{-19} \mathrm{C}$,
$\mathrm{r}=0.53 Å=0.53 \times 10^{-10} \mathrm{~m}$
By formula, potential energy,
$\mathrm{U}=\frac{\mathrm{l}}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}$
$=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{0.53 \times 10^{-10}}$
$=-\frac{144}{53}=-27.17 \mathrm{eV}$
(b) Kinetic energy of electron
$=+\frac{27.17}{2}=13.585 \mathrm{eV}$.
Total energy of electron $=-27.17+13.585=-13.585 \mathrm{eV}$.
When electron is free, energy becomes zero.
Then Work done $=$ increase in energy of electron $=0-(-13.585)=13.835 \mathrm{eV}$.
(c) Potential energy,
$\mathrm{U}=\frac{1}{4 \pi \varepsilon_0} . \mathrm{q}_1 \mathrm{q}_2\left(\frac{1}{\mathrm{r}_1}-\frac{1}{\mathrm{r}_2}\right)$ $=9 \times 10^9 \times\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)$ $\left(\frac{1}{0.53 \times 10^{-10}}-\frac{1}{1.06 \times 10^{-10}}\right) \mathrm{J}=-13.585 \mathrm{eV}$. Kinetic energy of electron $=+13.585 \mathrm{eV}$.
$\mathrm{q}_2=-1.6 \times 10^{-19} \mathrm{C}$,
$\mathrm{r}=0.53 Å=0.53 \times 10^{-10} \mathrm{~m}$
By formula, potential energy,
$\mathrm{U}=\frac{\mathrm{l}}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}}$
$=\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)}{0.53 \times 10^{-10}}$
$=-\frac{144}{53}=-27.17 \mathrm{eV}$
(b) Kinetic energy of electron
$=+\frac{27.17}{2}=13.585 \mathrm{eV}$.
Total energy of electron $=-27.17+13.585=-13.585 \mathrm{eV}$.
When electron is free, energy becomes zero.
Then Work done $=$ increase in energy of electron $=0-(-13.585)=13.835 \mathrm{eV}$.
(c) Potential energy,
$\mathrm{U}=\frac{1}{4 \pi \varepsilon_0} . \mathrm{q}_1 \mathrm{q}_2\left(\frac{1}{\mathrm{r}_1}-\frac{1}{\mathrm{r}_2}\right)$ $=9 \times 10^9 \times\left(1.6 \times 10^{-19}\right) \times\left(-1.6 \times 10^{-19}\right)$ $\left(\frac{1}{0.53 \times 10^{-10}}-\frac{1}{1.06 \times 10^{-10}}\right) \mathrm{J}=-13.585 \mathrm{eV}$. Kinetic energy of electron $=+13.585 \mathrm{eV}$.
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