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)?

n = 3 n = 4 n = 2 None of the above

n = 4

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