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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 $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\ N\,m^2/C^2$)
A
$2.1 \times 10^{-8}$ m
B
$2.1 \times 10^{-11}$ m
C
$2.1 \times 10^{-9}$ m
D
$2.1 \times 10^{-10}$ m
Detailed Solution
For an electron in a Bohr orbit, the magnitude of total energy $|E| = \frac{kq_1q_2}{2r} = \frac{ke^2}{2r}$
$3.4 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times r}$
$r = 2.1 \times 10^{-10}$ m