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 × 1

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 $= 1.6 \times 1 0^{- 19}$ J, $e = 1.6 \times 1 0^{- 19}$ C and $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ N m $^{2}$ /C $^{2}$ )NEET 2026 (Cancelled)

TestHub · Accepted Answer

The total energy of an electron in a hydrogen atom is given by $E = - \frac{1}{4 π ε _{0}} \frac{e ^{2}}{2 r}$

Substituting the given values:

$- 3.4 \times 1.6 \times 1 0^{- 19} = - \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{2 r}$

$r = \frac{9 \times 1 0 ^{9} \times 1.6 \times 1 0 ^{- 19}}{2 \times 3.4}$

$r = \frac{14.4 \times 1 0 ^{- 10}}{6.8}$

$r \approx 2.117 \times 1 0^{- 10}$ m

This is approximately $2.1 \times 1 0^{- 10}$ m.

Answer: $2.1 \times 1 0^{- 10}$ m

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