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 $^2$ )

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}

J,

e = 1.6 \times 10^{-19}

C and

\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}

N m

^2

/C

^2

)

Quantrex Academy · Accepted Answer

Correct answer: C. 2.1 × 10^ -10 m

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

In the first excited state of hydrogen atom, the energy of its...