Question

Q/C (a) Find the electric potential difference ΔVe required to stop an electron (called a "stopping potential") moving with an initial speed of 2.85 × 10⁷ m/s

Answer 1

The electric potential difference $\triangle V_e$ required to stop an electron (called a "stopping potential") moving with an initial speed (v) of 2.85 × 10⁷ m/s is $2.309\times10^3V$.

How to calculate the electric potential difference?

We know, that the kinetic energy is:

$K=qV_0=\frac{1}{2} mv^2$

where,

q is the charge

$q=1.6\times10^{-19}C$

$V_0$ is the stopping potential

m is the mass of the electron

$m=9.1\times10^{-31}kg$

v is the speed of the electron

Now,

$V_0=\frac{1}{2} \frac{9.1\times10^{-31}\times(2.85\times10^7)^2}{1.6\times10^{-19}} =2.309\times10^3V$

Hence, the electric potential difference is $2.309\times10^3 V$.

To learn more about total electric potential, refer to:

brainly.com/question/16843018

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