It is found experimentally that $13.6 \; eV$ energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
The total energy of the electron in the ground state of a hydrogen atom is $E = -13.6 \; eV = -13.6 \times 1.6 \times 10^{-19} \; J = -2.176 \times 10^{-18} \; J \approx -2.2 \times 10^{-18} \; J$.
Using the formula for total energy $E = -\frac{e^2}{8 \pi \varepsilon_0 r}$,we can solve for the orbital radius $r$:
$r = -\frac{e^2}{8 \pi \varepsilon_0 E} = -\frac{(9 \times 10^9 \; N \cdot m^2/C^2) \times (1.6 \times 10^{-19} \; C)^2}{2 \times (-2.176 \times 10^{-18} \; J)} \approx 5.3 \times 10^{-11} \; m$.
For the velocity $v$,we use the relation $v = \sqrt{\frac{e^2}{4 \pi \varepsilon_0 m r}}$:
$v = \sqrt{\frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31}) \times (5.3 \times 10^{-11})}} \approx 2.2 \times 10^6 \; m/s$.
Using the formula for total energy $E = -\frac{e^2}{8 \pi \varepsilon_0 r}$,we can solve for the orbital radius $r$:
$r = -\frac{e^2}{8 \pi \varepsilon_0 E} = -\frac{(9 \times 10^9 \; N \cdot m^2/C^2) \times (1.6 \times 10^{-19} \; C)^2}{2 \times (-2.176 \times 10^{-18} \; J)} \approx 5.3 \times 10^{-11} \; m$.
For the velocity $v$,we use the relation $v = \sqrt{\frac{e^2}{4 \pi \varepsilon_0 m r}}$:
$v = \sqrt{\frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(9.1 \times 10^{-31}) \times (5.3 \times 10^{-11})}} \approx 2.2 \times 10^6 \; m/s$.
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