Question

An electron (m=9.11x10^-31 kg) moves in a circle whose radius is 2.00x10^-2 m. If the force acting on the electron is 4.60x10^-14 N, what is it's speed?

Answer 1

Answer:

The speed of the electron is $3.178\times 10^7\ m/s$.

Explanation:

Given:

Mass of electron is, $m=9.11\times 10^{-31}\ kg$

Radius of circle is, $R=2.00\times 10^{-2}\ m$

Force acting on the electron is, $F=4.60\times 10^{-14}\ N$

Now, we know that, for a circular turn, the force acting on the electron is due to centripetal force. Centripetal force acting on the electron is given as:

$F=\frac{mv^2}{R}$

Here, 'v' is the velocity of the electron.

Now, plug in all the given values and solve for 'v'. This gives,

$4.60\times 10^{-14}=\frac{9.11\times 10^{-31}v^2}{2.00\times 10^{-2}}\\\\9.11\times 10^{-31}v^2=4.60\times 10^{-14}\times 2.00\times 10^{-2}\\\\9.11\times 10^{-31}v^2=9.2\times 10^{-16}\\\\v^2=\frac{9.2\times 10^{-16}}{9.11\times 10^{-31}}\\\\v^2=1.01\times 10^{15}\\\\v=\sqrt{1.01\times 10^{15}}\\\\v=3.178\times 10^7\ m/s$

Therefore, the speed of the electron is $3.178\times 10^7\ m/s$.