The electrostatic potential energy of the electron in an orbit of hydrogen is -6.8 eV . The speed of the…

The electrostatic potential energy of the electron in an orbit of hydrogen is -6.8 eV . The speed of the electron in this orbit is ( C is speed of light in vacuum)
  1. $\frac{C}{137}$
  2. $\frac{C}{274}$
  3. $\frac{2 C}{137}$
  4. $\frac{3 C}{137}$

Solution

$\mathrm{U}=-6.8 \mathrm{eV}$
$\therefore$ Total energy, $\mathrm{E}_{\mathrm{n}}=\frac{\mathrm{U}}{2}=\frac{-6.8}{2}=-3.4 \mathrm{eV}$
$\Rightarrow \frac{-13.6}{\mathrm{n}^2}=-3.4$ $\therefore \quad \mathrm{n}=2$ $\therefore$ Speed, $\mathrm{v}_{\mathrm{n}}=\left(\frac{\mathrm{C}}{137}\right) \cdot \frac{1}{\mathrm{n}}$
for $\mathrm{n}=2, \mathrm{v}_2=\left(\frac{\mathrm{C}}{137}\right) \times \frac{1}{2}=\frac{\mathrm{C}}{274}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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