What is the speed of a 680 eV electron?
What is the speed of a 3.50 keV electron?
(1/2) m v^2 = 680 eV * 1.602*10^-19 Joules/eV
(1/2)(9.1*10^-31) v^2 = 680*1.6*10-19
v^2 =680*1.6*10-19 / (4.55*10^-31)
To calculate the speed of an electron with a given energy, we will use the equation for kinetic energy:
KE = (1/2) mv^2
Where KE is the kinetic energy, m is the mass of the electron, and v is the speed of the electron.
First, we need to convert the given energy values from electron volts (eV) to joules (J) by using the conversion factor:
1 eV = 1.6 × 10^-19 J
For the first question, an electron with an energy of 680 eV:
680 eV * (1.6 × 10^-19 J/eV) = 1.088 × 10^-16 J
Next, we need to find the mass of an electron. The mass of an electron is approximately 9.1 × 10^-31 kg.
Now, we can rearrange the kinetic energy equation to solve for velocity:
v^2 = (2KE) / m
v^2 = (2 * (1.088 × 10^-16 J)) / (9.1 × 10^-31 kg)
v^2 = 2.39 × 10^14 m^2/s^2
Taking the square root of both sides, we can find the speed of the electron:
v ≈ 1.545 × 10^7 m/s
Therefore, the speed of a 680 eV electron is approximately 1.545 × 10^7 m/s.
For the second question, an electron with an energy of 3.50 keV:
3.50 keV * (1.6 × 10^-19 J/eV) = 5.60 × 10^-17 J
Using the same mass of an electron:
v^2 = (2 * (5.60 × 10^-17 J)) / (9.1 × 10^-31 kg)
v^2 ≈ 1.23 × 10^15 m^2/s^2
Taking the square root:
v ≈ 1.110 × 10^7 m/s
Therefore, the speed of a 3.50 keV electron is approximately 1.110 × 10^7 m/s.