What is the kinetic energy of an electron {mass = 9.1 x 10^ - 31 kg} when it is moving at 3.0 x 10^7 m/s {1/10 the velocity of light}?
KE = 1/2 m v^2
KE = 1/2 (9.1 x 10^ - 31)(3.0 x 10^7)
KE = 1.365 x 10^ -23
I don't know if this is right because what got me confused is the 1/10 the velocity of light.
No. You did not square velocity.
To find the kinetic energy of an electron, you can use the formula KE = 1/2 m v^2, where KE is the kinetic energy, m is the mass of the electron, and v is its velocity.
In this case, the mass of the electron is given as 9.1 x 10^-31 kg and the velocity is given as 3.0 x 10^7 m/s. However, you mentioned that the velocity is "1/10 the velocity of light," which is a significant clue.
The velocity of light is approximately 3.0 x 10^8 m/s. Therefore, if the electron's velocity is 1/10 the velocity of light, it would be 0.1 x 3.0 x 10^8 = 3.0 x 10^7 m/s.
Now, plugging in the values into the formula:
KE = 1/2 (9.1 x 10^-31 kg) (3.0 x 10^7 m/s)^2
Evaluating the expression:
KE = 1/2 (9.1 x 10^-31 kg) (9.0 x 10^14 m^2/s^2)
KE = 4.095 x 10^-17 J
Therefore, the kinetic energy of the electron is 4.095 x 10^-17 Joules.
Please note that the units for the kinetic energy are Joules (J).