Answer to Question #84139 in Physics for Yoseph Zenaw

Question

An electronic starts from rest and falls through a potential rise of 80 v. What is its final speed?

Answer 1

We can find the final speed of the electron from the work-kinetic energy theorem. It states that the work done by potential difference is equal to the change in kinetic energy of the electron:

$$∆PE=KE_f - KE_i,$$ $$qV = \frac{1}{2}mv_f^2 - 0,$$

here,

$$q$$

is the charge of the electron,

$$V$$

is the potential difference,

$$m$$

is the mass of the electron and

$$v_f$$

is the final speed of the electron.

Then, from this formula we can find the final speed of the electron:

$$v_f = \sqrt{\frac{2qV}{m}}.$$

Let's substitute the numbers:

$$v_f = \sqrt{\frac{2 \cdot 1.6 \cdot 10^{-19} C \cdot 80 V}{9.11 \cdot 10^{-31} kg}} =5.3 \cdot 10^6 \frac{m}{s}.$$

Answer:

$$v_f = 5.3 \cdot 10^6 \frac{m}{s}$$

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