Q43 (2024)
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 $1eV = 1.6 \times 10^{-19}$ J, $e = 1.6 \times 10^{-19}$ C and $\frac{1}{4\pi\varepsilon_{0}} = 9 \times 10^{9}$ N m $^{2}$ /C $^{2}$ )
- $2.1 \times 10^{-8} \text{ m}$
- $2.1 \times 10^{-11} \text{ m}$
- $2.1 \times 10^{-9} \text{ m}$
- $2.1 \times 10^{-10} \text{ m}$
### Related Formula
$$|E| = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{2r}$$
### Core Logic
$$\frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 3.4 \times 1.6 \times 10^{-19}} = r$$
$$r = \frac{9 \times 1.6 \times 10^{-10}}{6.8} \approx 2.1176 \times 10^{-10} \text{ m} \approx 2.1 \times 10^{-10} \text{ m}$$
### Pattern Recognition
Bohr orbit radius relation to total binding energy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
