Question

M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.

Answer 1

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is $6.49\times10^6m/s$

How to calculate the speed of the electron?

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

$\triangle K+\triangle$ U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

$(\frac{1}{2} mv^2)+(-q\triangle V)$ =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let $v_f$ and $v_i$ represent the final and initial speed.

Here, $v_i$ =0

Solving for $v_f$, we get:

$v_f=\sqrt{\frac{-2q\triangle V}{m} }$

$=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }$

=6.49$\times10^6$m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

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