An electron of mass 9.1 x 10{}^{-31} kg moves with a speed of 2.0 x 10^6 ms{}^{-1} round the nucleus of an atom in a Circular path of radius

Correct Answer is : 6.0 x 10−8{}^{-8}−8N

Correct Answer is : 6.0 x 10−8{}^{-8}−8N

Test Index

WAEC Physics 2019 Paper

Question : 31

Total: 49

{}^{-31}

^6

{}^{-1}

Solution:

Concept:

An electron moving in a circular path experiences a centripetal force directed toward the nucleus.

This force is given by

F = \frac{mv^2}{r}

Explanation:

Given mass

m = 9.1 \times 10^{-31}\ \mathrm{kg}

, speed

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

, and radius

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

Substitute into

F = \frac{mv^2}{r}

F = \frac{9.1 \times 10^{-31} \times (2.0 \times 10^6)^2}{6.1 \times 10^{-11}}

F = \frac{9.1 \times 10^{-31} \times 4.0 \times 10^{12}}{6.1 \times 10^{-11}}

F = \frac{36.4 \times 10^{-19}}{6.1 \times 10^{-11}}

F \approx 5.97 \times 10^{-8}\ \mathrm{N}

F \approx 6.0 \times 10^{-8}\ \mathrm{N}

Answer:

The correct option is B.

6.0 \times 10^{-8}\ \mathrm{N}

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