Question

An electron with an initial speed of 660,000 m/s is brought to rest by an electric field. What was the initial kinetic energy of the electron, in electron volts?

Answer 1

Answer:

Kinetic energy of the electron $E_k=1.23\ eV$

Explanation:

It is given that,

Initial speed of an electron, u = 660000 m/s

Final speed of the electron, v = 0 (at rest)

The kinetic energy of an electron is possessed due to the motion of an electron. The mathematical formula for the kinetic energy is given by :

$E_k=\dfrac{1}{2}mv^2$

m is the mass of electron

$E_k=\dfrac{1}{2}\times 9.1\times 10^{-31}\times (660000)^2$

$E_k=1.98\times 10^{-19}\ J$

Since, $1\ eV=1.602\times 10^{-19}\ J$

$E_k=\dfrac{1.98\times 10^{-19}}{1.602\times 10^{-19}}$

$E_k=1.23\ eV$

So, the initial kinetic energy of the electron is 1.23 eV. Hence, this is the required solution.