The photoelectric work function of a metal is the minimum energy needed to eject an electron by irradiating the metal with light. For calcium, this work function equals 4.34 x 10-19 J. If light with a wavelength of 256 nm impinges on a piece of calcium, what is the speed of the ejected electron in m/s?
I was able to find the kinetic energy of the electron as 3.4x10^-19J but I can't figure out how to convert that to m/s
K.E. = 1/2 m v^2
look up the mass of the electron (in kg), and solve for v
To convert the kinetic energy of an electron to its speed, we can use the following formula:
Kinetic Energy = (1/2) * mv^2,
where m is the mass of the electron and v is its speed. We know the kinetic energy of the electron to be 3.4 x 10^-19 J.
First, let's determine the mass of an electron. The mass of an electron is approximately 9.10938356 x 10^-31 kg.
Now we can substitute the values into the formula and solve for v:
3.4 x 10^-19 J = (1/2) * (9.10938356 x 10^-31 kg) * v^2
To isolate v^2, divide both sides of the equation by (1/2) * (9.10938356 x 10^-31 kg):
v^2 = (2 * 3.4 x 10^-19 J) / (9.10938356 x 10^-31 kg)
v^2 = 7.46196165 x 10^11 m^2/s^2
To find v, take the square root of both sides of the equation:
v = √(7.46196165 x 10^11 m^2/s^2)
v ≈ 2.73 x 10^5 m/s
Therefore, the speed of the ejected electron is approximately 2.73 x 10^5 m/s.