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.

Answer 2

Acceleration is defined as the rate of change of electron velocity in an orbit. The magnitude of the electron's acceleration is 3.95×10¹³ m/sec².While the kinetic energy of the electron is 4.54×10⁻¹⁸.

What is the kinetic energy of electrons?

The kinetic energy (KE) of electrons is defined as the product of one-half of the mass of the electron to the square of the velocity at which electrons revolve in orbit.

The given data in the problem will be

V is the voltage at the end of wire=18 cm

l is the length of wire =8 cm

q is the charge of electron=1.6 x 10⁻⁸ c

m is the mass of electron=9.11 x 10⁻³¹ Kg

v is the velocity of electon=?

KE is the kinetic energy of electron=?

The electric force is given by the product of charge and voltage per unit length. While the mechanical force is given by the product of mass and acceleration.

$\rm F=qE\\\\\rm E=\frac{V}{l} \\\\\rm F=ma\\\\ma=\frac{qV}{l}\\\\ \rm a=\frac{qV}{Lm} \\\\\rm a=\frac{1.6\times10^{-19}\times18}{0.08\times9.11\times10^{-31}} \\\\\rm a=3.95\times10{13}\;m/sec^2$

Hence the magnitude of the electron's acceleration is 3.95×10¹³.

According to Newton's third equation of motion

since the initial velocity of electron u=0 so

$\rm v^2=u^2+2as\\\\\rm v^2=2as\\\\v=\sqrt{2as} \\\\ \rm v=\sqrt{2\times3.95\times10^{31}\times4\times10^{-18}}\\\\\rm v=3160000\\\\\\\\E=\frac{1}{2}mv^2 \\\\ \rm E=\frac{1}{2}\times9.11\times10^{-31}(3160000)^2\\\\\rm E=4.54\times10^{-18} \;J$

Hence the kinetic energy of the electron is 4.54×10⁻¹⁸.

To learn more about kinetic energy refer to the link ;

brainly.com/question/24134093