Question
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.
✓
Answer
Total energy of the electron in hydrogen atom is $-13.6 eV =-13.6 \times 1.6 \times 10^{-19} J$
$=-2.2 \times 10^{-18} J$
Thus from Eq., we have
$-\frac{e^2}{8 \pi \varepsilon_0 r}= E$
This gives the orbital radius
$r=-\frac{e^2}{8 \pi \varepsilon_0 E}$
$=-\frac{\left(9 \times 10^9 Nm ^2 / C ^2\right)\left(1.6 \times 10^{-19} C \right)^2}{(2)\left(-2.2 \times 10^{-18} J\right)}$
$=5.3 \times 10^{-11} m$
The velocity of the revolving electron can be computed from Eq. with $m =9.1 \times 10^{-31} \ kg$,
$\frac{1}{2} m v^2=\frac{e^2}{4 \pi \epsilon_0 r^2}$ thus velocity of electron is given by : $-$
$v=\frac{e}{\sqrt{4 \pi \varepsilon_0 m r^2}}$
$=2.2 \times 10^6 m / s$
$=-2.2 \times 10^{-18} J$
Thus from Eq., we have
$-\frac{e^2}{8 \pi \varepsilon_0 r}= E$
This gives the orbital radius
$r=-\frac{e^2}{8 \pi \varepsilon_0 E}$
$=-\frac{\left(9 \times 10^9 Nm ^2 / C ^2\right)\left(1.6 \times 10^{-19} C \right)^2}{(2)\left(-2.2 \times 10^{-18} J\right)}$
$=5.3 \times 10^{-11} m$
The velocity of the revolving electron can be computed from Eq. with $m =9.1 \times 10^{-31} \ kg$,
$\frac{1}{2} m v^2=\frac{e^2}{4 \pi \epsilon_0 r^2}$ thus velocity of electron is given by : $-$
$v=\frac{e}{\sqrt{4 \pi \varepsilon_0 m r^2}}$
$=2.2 \times 10^6 m / s$
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