An electron possessing a kinetic energy of 100 eV circulates in a...

Kinetic energy of the electron, $K = 100 \text{ eV} = 100 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-17} \text{ J}$

The radius of the circular path is given by $R = \dfrac{\sqrt{2mK}}{qB}$

Rearranging for magnetic field $B$:

$B = \dfrac{\sqrt{2mK}}{qR}$

Substituting the standard values ($m = 9.1 \times 10^{-31} \text{ kg}$, $q = 1.6 \times 10^{-19} \text{ C}$, $R = 0.1 \text{ m}$):

$B = \dfrac{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}}{1.6 \times 10^{-19} \times 0.1}$

$B = \dfrac{\sqrt{29.12 \times 10^{-48}}}{1.6 \times 10^{-20}} = \dfrac{5.396 \times 10^{-24}}{1.6 \times 10^{-20}} \approx 3.37 \times 10^{-4} \text{ T}$

The number of revolutions per second is the frequency $f$, given by:

$f = \dfrac{qB}{2\pi m}$

$f = \dfrac{1.6 \times 10^{-19} \times 3.37 \times 10^{-4}}{2 \times 3.14 \times 9.1 \times 10^{-31}}$

$f = \dfrac{5.392 \times 10^{-23}}{57.148 \times 10^{-31}} \approx 9.43 \times 10^6 \text{ Hz}$

Rounding off to two significant figures, we get $B = 3.4 \times 10^{-4} \text{ T}$ and $f = 9.4 \times 10^6 \text{ Hz}$.