An electron has an initial speed of 5.40×10 6 m/s. What potential difference (si

Question

An electron has an initial speed of 5.40×106 m/s. What potential difference (sign and magnitude) is required to bring the electron to rest?

ANSWER IS NOT 83VOLTS

What potential difference is required to reduce the kinetic energy of the electron by a factor of 3 compared with the initial value?

What potential difference is required to reduce the speed of the electron by a factor of 3 compared with the initial speed?

PLEASE GO ABOUT THIS DIFFERENTLY, Ive posted this question 4 times now and everyone keeps getting 83V for the first one and its wrong, you do something else. I really need help.
Hints given:You shouldn't be dividing the initial speed by 2 at all, other than dividing the square of the initial speed times mass by two, as per the kinetic energy formula. Once you have found that, it should be easy to find potential energy. The only time you directly touch initial speed is when you divide it by 3 before squaring it in part 3.

Explanation / Answer

The potential difference is V = KE / q = (1/2)mv^2 / q = mv^2 / 2q

        V = (9.1*10^-31 kg)(5.4*10^6 m/s)^2 / 2*1.6*10^-19 = - 82.9 V

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