Answer
(a) $n = 31$
(b) $v_{31} = 7.06 \times 10^4~m/s$
$E_{31} = -0.439~eV$
Work Step by Step
(a) We can find the quantum number:
$r_n = n^2~a_B = 50~nm$
$n^2 = \frac{50~nm}{a_B}$
$n = \sqrt{\frac{50~nm}{a_B}}$
$n = \sqrt{\frac{50~nm}{0.0529}}$
$n = 31$
(b) We can find the speed:
$v_n = \frac{v_1}{n}$
$v_{31} = \frac{2.19\times 10^6~m/s}{31}$
$v_{31} = 7.06 \times 10^4~m/s$
We can find the energy:
$E_n = -\frac{13.6~eV}{n^2}$
$E_{31} = -\frac{13.6~eV}{31^2}$
$E_{31} = -0.439~eV$
