Question

Find the kinetic energy ofan electron that moves at half the speed of light to four significant digits.

Answer 1

Answer:

Ke=electron kinetic energy=$10.24*10^{33} J$

Explanation:

The electron has a mass of $9.1*10^{-31} kg$

The speed of light in a vacuum is a universal constant with the value 299 792 458 m / s (186 282,397 miles / s), although it is usually close to $3*10^{8} \frac{m}{s}$

Kinetic energy (K) is the energy associated with bodies that are in motion, depends on the mass and speed of the body and is calculated using the formula:

$K=\frac{1}{2} *m*v^{2}$        Equation(1)

K=kinetic energy (J)

m =mass of the body (kg)

v= speed of the body$(\frac{m}{s} )$

for this problem We replace in the equation (1)

$me=9.1*10^{-31} kg$ = electron mass

$ve=1.5*10^{8} \frac{m}{s}$=Half the speed of light

=electron speed

We replace in  the  equation (1) :

$Ke=\frac{1}{2}*me*ve^{2}$

$Ke=\frac{1}{2} *9.1*10^{-31} *(1.5*10^{8} )^{2}$

$Ke=4.55*10^{-31} *(1.5)^{2} *(10^{8} )^{2}$

$Ke=10.24*10^{33} J$

The energy kinetic of the electron is $10.24*10^{33} Joules$