Answer
a) $6.62\times 10^{-25}Kg.\frac{m}{s}$
b) $3.17\times 10^{-34}J.s$
Work Step by Step
(a) We can find the linear momentum of an electron as
$P_n=mv_n=m(\frac{2\pi Ke^2}{nh})$
We plug in the known values to obtain:
$P_3=9.11\times 10^{-31}(\frac{2\pi(8.99\times 10^9)(1.60\times 10^{-19})^2}{3(6.634\times 10^{-34})})$
$P_3=6.62\times 10^{-25}Kg.\frac{m}{s}$
(b) The angular momentum of an electron can be determined as
$L_n=\frac{hn}{2\pi}$
We plug in the known values to obtain:
$L_3=\frac{3h}{2\pi}$
$L_3=\frac{3(6.63\times 10^{-34})}{2\pi}$
$L_3=3.17\times 10^{-34}J.s$
