An atom of hydrogen has an electron in some excited state having radius of 8.48 × 10-10 m. The electron will lie in orbit of which principle quantum number (n) (given Bohr radius, a0 = 5.3 × 10-11 m)?
Answer & explanation
Correct answer: option 2
rn = n2h2ε0/mπe2 or n2r1 (r1 is Bohr radius, a0)
n2 = rn/a0
$n^2= \frac{8.48 \times 10^{-10}}{5.3 \times 10^{-11}}= 16$
n = 4