In summary, the conversation discusses the relationship between cutoff potential and kinetic energy measured in electron volts. The student is attempting to solve a physics lab question using the equation KE=qV, where q is the charge on an electron and V is the cutoff potential. However, there are some errors in the student's understanding, such as the incorrect charge of an electron and confusion between current and cutoff potential. The conversation also addresses the possibility of negative kinetic energy and the importance of understanding the concepts behind formulas.
in my class we are doing this physics lab where we get the data, so i have used the equation KE=qV where q is 1.60*10^-19 C which is the charge on an electron, and V is the current at the cutoff potential. So for yellow light i have the values 0.25 J/C. Now I got an answer of 4.0*10^18 J when solving the equation. However, the question also says write your answer in electron volts, I know that one eV is 1.60*10^-19 J/eV, same value as the charge on an electron but different units. anyways i divided 4.0*10^18 J by 1.60*10^-19 J/eV and i got 0.25 eV same as what i had before. I am so confused. This screenshot might make better sense.
I have made all of my corrections to the charge of an electron and the amount of joules in an electron volt. Now i am trying to solve for the max KE of the electron using the equation KE=qV, where q is the charge on an electron: -1.60*10^-19 C, and V is the cutoff potential when the current is 0. Using this information for green light which has a cutoff potential of 0.25 V. I can now solve for the maximum KE possible, when i do this i get an answer of -4.0*10^-20 J. This question also wants me to convert J to electron volts, since i know that one electron volt is 1.60*10^-19 J/eV, i divided -4.0*10^-20 J by 1.60*10^-19 J/eV to get an answer of -0.25 eV.
Instead of just using formulas, try to see what you're really doing. You're firing electrons with an initial k.e. of 0.25 ev at a NEGATIVE plate. So what happens? As you inrease the voltage NEGATIVELY, higher and higher REPULSIVE force is applied to the speeding electrons until the voltage is negative enough to apply just enough repulsive force to prevent the electrons from arriving at the plate. They instead do a 180 in front of the plate and head back to the emitter.
So you have an initial k.e. of 0.25 ev but the electrons LOSE energy as they move to the plate. That energy is qV with q negative and V negative.
Don't just depend on equations to give you the right polarity in any given problem. Think it through.
And no, there is no such thing as negative kinetic energy. k.e. = 1/2 mv2 so even if v is negative, k.e. is positive.