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NEET · Physics · STD 12 - 12. atoms

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 \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 N m ^2 / C ^2\) )

  1. A \(2.1 \times 10^{-9} m\)
  2. B \(2.1 \times 10^{-8} m\)
  3. C \(2.1 \times 10^{-10} m\)
  4. D \(2.1 \times 10^{-11} m\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2.1 \times 10^{-10} m\)

Step-by-step Solution

Detailed explanation

(C) \(2.1 \times 10^{-10} m\)
The total energy of an electron in a hydrogen atom is given by \(E=-\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{2 r}\)
Substituting the given values:
\(-3.4 \times 1.6 \times 10^{-19}=-\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{2 r}\)
\(r=\frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 3.4}\)
\(r=\frac{14.4 \times 10^{-10}}{6.8}\)
\(r \approx 2.117 \times 10^{-10} m\)