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WAEC Physics 2019 Paper

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Question : 30

Total: 49

An electron of mass

9.1 \times 10^{-31}

kg moves with a speed of

2.0 \times 10^{6}

^{-1}

round the nucleus of an atom in a Circular path of radius

6.1 \times 10^{11}

m. Determine the angular speed of the electron.

3.28 x 10 $^{16}$ rad s $^{-1}$
8.55 x 10 $^{3}$ rad s $^{-1}$
9.11 x 10 $^{13}$ rad s $^{-1}$
5.22 x 10 $^{15}$ rad s $^{-1}$

Solution:

Concept:

Angular speed

\omega

is the rate of rotation around the nucleus, given by

\omega = \frac{v}{r}

, where

v

is linear speed and

r

is the radius of the circular path.

Explanation:

Given:

v = 2.0 \times 10^{6}\ \mathrm{m\,s^{-1}}

and

r = 6.1 \times 10^{-11}\ \mathrm{m}

.

The mass of the electron is not needed in this calculation.

Substitute into

\omega = \frac{v}{r}

:

\omega = \frac{2.0 \times 10^{6}}{6.1 \times 10^{-11}}

= \frac{2.0}{6.1} \times 10^{6 - (-11)}

= 0.328 \times 10^{17}

= 3.28 \times 10^{16}\ \mathrm{rad\,s^{-1}}

.

Answer:

The angular speed of the electron is

3.28 \times 10^{16}\ \mathrm{rad\,s^{-1}}

, so the correct option is A.

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