Answer
(a) $1.52\times 10^{-16}s$
(b) $0.00105A$
Work Step by Step
(a) We know that
$v_1=\frac{2\pi ke^2}{1.h}$
We plug in the known values to obtain:
$v_1=\frac{(2)(3.14)(9\times 10^9)(1.6\times 10^{-19})^2}{6.63\times 10^{-34}}$
$v_1=21.82\times 10^5m/s$
Now $t=\frac{(2)(3.14)(5.29\times 10^{11})}{21.82\times 10^5}$
$t=1.52\times 10^{-16}s$
(b) The required current can be determined as
$i=\frac{e}{t}$
We plug in the known values to obtain:
$i=\frac{1.6\times 10^{-19}}{1.52\times 10^{-16}}$
$i=1.05\times 10^{-3}A=0.00105A$
