Answer
(a) $\omega=4.12\times 10^{16} rad/s$
(b) $6.56\times 10^{15}$ orbits
(c) $a=8.98\times 10^{22}m/s^2$
Work Step by Step
(a) We know that
$\omega=\frac{v}{r}$
We plug in the known values to obtain:
$\omega=\frac{2.18\times 10^6m/s}{5.29\times 10^{-11}m}$
$\omega=4.12\times 10^{16}rad/s$
(b) We can find the angular velocity in rev/s as follows:
$\omega=\frac{4.12\times 10^{16}rad/s}{2\pi rad/rev}$
$\omega=6.56\times 10^{15}rev/s$
Thus, every second the electron makes $6.56\times 10^{15}$ revolutions or orbits about the proton.
(c) The centripetal acceleration can be determined as follows:
$a_{cp}=\frac{v^2}{r}$
We plug in the known values to obtain:
$a_{cp}=\frac{(2.18\times 10^6m/s)^2}{5.29\times 10^{-11}m}$
$a_{cp}=8.98\times 10^{22}m/s^2$
