In the second Bohr orbit of $He X^{+}$ , find the time for one revolution of the electron in units of $1 0^{- 16} s$ , to two decimals. Use $r_{n} = 0.529 \times 1 0^{- 10} n^{2} / Z m$ and $v = 2.18 \times 1 0^{6} Z / n m s^{- 1}$ .

Numerical answer type, answer in x 10^-16 s, tolerance 0.03.

Answer: 3.05 x 10^-16 s

Z = 2, n = 2: r = 0.529e-10 x 4/2 = 1.0580e-10 m and v = 2.18e6 x 2/2 = 2.180e+06 m/s. T = 2 pi r/v = 3.0494e-16 s = 3.05 x

1 0^{- 16}