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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=1.6×10⁻¹⁹ J, e=1.6×10⁻¹⁹ C and 1/(4πε₀)=9×10⁹ N m² C⁻²)

Asked in NEET 2026 · Energy levels and ionisation

Answer: (2) 2.1×10⁻¹⁰ m

Step-by-step solution

In a Bohr orbit the total energy is E=-(ke²)/(2r).

r=(ke²)/(2|E|)=(9×10⁹×(1.6×10⁻¹⁹)²)/(2×3.4×1.6×10⁻¹⁹).

r=(9×10⁹×1.6×10⁻¹⁹)/(6.8)=2.12×10⁻¹⁰ m.

Check: n=2 gives 4×0.53 Å =2.12 Å.

Why the other options are wrong

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