Question

An electron is to be accelerated from a velocity of 5.00×106 m/s to a velocity of 9.90×106 m/s . through what potential difference must the electron pass to accomplish this

Answer 1

Potential difference needed= 207.9 V

Explanation=

using work energy principle

Work done on the electron= change in kinetic energy

W= 1/2m Vf²- 1/2 mVi²

m= mass of electron=9.11 x 10⁻³¹ Kg

Vf= final velocity= 9.9 x 10⁶ m/s

Vi= initial velocity= 5 x 10⁶ m/s

so W = 1/2 (9.11 x 10⁻³¹ ) (9.9 x 10⁶ )²- 1/2 (9.11 x 10⁻³¹ ) (5 x 10⁶ )²

W=3.33 x 10⁻¹⁷ J

Now work done= q V

q= charge of electron= 1.6 x 10⁻¹⁹J

V= potential difference

3.33 x 10⁻¹⁷= (1.6 x 10⁻¹⁹) V

v= 207.9 V

Answer 2

The potential difference accomplished by the electron during its passage is 207.9 Volts.

What is Work-Energy Principle?

The work-energy principle says that "The work done due to the applied force is equal to the change in kinetic energy of the particle".

Given data:

The initial velocity of the electron is, $u = 5.00 \times 10^{6} \;\rm m/s$.

The final velocity of the electron is, $v = 9.90 \times 10^{6} \;\rm m/s$.

The mathematical expression for the work-energy principle is given as,

$W = \Delta KE\\\\W = \dfrac{1}{2}m(v^{2}-u^{2})$

Here, m is the mass of the electron.

Solving as,

$W = \dfrac{1}{2} \times 9.31 \times 10^{-31} \times ((9.90 \times 10^6)^2-(5.00 \times 10^6)^2)\\\\W =3.33 \times 10^{-17} \;\rm J$

The work done on the electron can also be expressed as,

$W = e \times V$

here,

e is the charge on the electron.

V is the potential difference.

Solving as,

$3.33 \times 10^{-17} = (1.6 \times 10^{-19}) \times V\\\\V = 207.9 \;\rm Volts$

Thus, we can conclude that the potential difference accomplished by the electron during its passage is 207.9 Volts.

Learn more about the electric potential here:

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