Q117 / 180 · NEET 2026 (Cancelled) · Physics
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 $= 1.6 \times 10^{-19}$ J, $e = 1.6 \times 10^{-19}$ C and $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ N m$^2$/C$^2$)
(Take $1$ eV $= 1.6 \times 10^{-19}$ J, $e = 1.6 \times 10^{-19}$ C and $\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ N m$^2$/C$^2$)
- A$2.1 \times 10^{-9}$ m
- B$2.1 \times 10^{-8}$ m
- Correct answerC$2.1 \times 10^{-10}$ m
- D$2.1 \times 10^{-11}$ m
Solution
The total energy of an electron in a hydrogen atom is given by $E = -\dfrac{1}{4\pi\varepsilon_0} \dfrac{e^2}{2r}$
Substituting the given values:
$-3.4 \times 1.6 \times 10^{-19} = -\dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2r}$
$r = \dfrac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 3.4}$
$r = \dfrac{14.4 \times 10^{-10}}{6.8}$
$r \approx 2.117 \times 10^{-10}$ m
This is approximately $2.1 \times 10^{-10}$ m.
Answer: $2.1 \times 10^{-10}$ m
Substituting the given values:
$-3.4 \times 1.6 \times 10^{-19} = -\dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2r}$
$r = \dfrac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 3.4}$
$r = \dfrac{14.4 \times 10^{-10}}{6.8}$
$r \approx 2.117 \times 10^{-10}$ m
This is approximately $2.1 \times 10^{-10}$ m.
Answer: $2.1 \times 10^{-10}$ m
Hence, the correct answer is option: 'C'.