Question

A voltage of 18 V is applied across the ends of a piece of copper wire 8 cm long. The mass of an electron is 9.11 x 10kg and its charge is 1.6 x 10-B c What is the magnitude of the electron's acceleration? Answer in units of m/s part 2 of 2 points After traveling 4 × 10-8 m, if it does not collide with a copper ion over this distance what is the kinetic energy of the electron

Answer 1

Explanation:

It is given that,

Voltage, V = 18 V

Length of the copper wire, l = 8 cm = 0.08 m

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

Charge on electron, $q=1.6\times 10^{-19}\ C$

Electric force is given by, $F=qE=\dfrac{qV}{l}$

Also, $ma=\dfrac{qV}{l}$

$a=\dfrac{qV}{lm}$

$a=\dfrac{1.6\times 10^{-19}\times 18}{0.08\times 9.11\times 10^{-31}}$

$a=3.95\times 10^{13}\ m/s^2$

Since, the electron is at rest initially, u = 0

$v^2=2as$, $s=4\times 10^{-8}\ m$

$v^2=2\times 3.95\times 10^{13}\times 4\times 10^{-8}$

$v^2=3160000$

The kinetic energy of the electron is :

$E=\dfrac{1}{2}\times m\times v^2$

$E=\dfrac{1}{2}\times 9.11\times 10^{-31}\times (3160000)^2$

$E=4.54\times 10^{-18}\ J$

Hence, this is the required solution.