Atomic StructureMedium

Question

In the first excited state of hydrogen atom, the energy of its electron is -3.4 eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately:

(Take 1 eV $= 1.6 \times 10^{- 19}\text{ }J,e = 1.6 \times 10^{- 19}C$ and $\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}\text{ }N{\text{ }m}^{2}/C^{2}$ )

Options

A.$2.1 \times 10^{- 9}\text{ }m$
B.$2.1 \times 10^{- 8}\text{ }m$
C.$2.1 \times 10^{- 10}\text{ }m$
D.$2.1 \times 10^{- 11}\text{ }m$

Solution

Sol. [Key Concept: Bohr model — orbital radius r ∝ n²/Z] $\frac{KQ^{2}}{2r} = 3.4eV$

$$\begin{matrix} & \frac{9 \times 10^{9} \times \left( 1.6 \times 10^{- 19} \right)^{2}}{2 \times 3.4 \times 1.6 \times 10^{- 19}} = r \\ & 2.1176 \times 10^{- 10} = r \\ & r = 2.1 \times 10^{- 10}\text{ }m \end{matrix}$$

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