Question

Two electrodes are maintained at a potential differences of 50V. An electron moving from cathode to anode gains kinetic energy equal to

Answer 1

The answer to the question is $80.1\times 10^-19\ Joule$ i.e the kinetic energy of the electron is $80.1\times 10^-19\ Joule$

CALCULATION:

As per the question, the potential difference between electrodes V = 50 V.

The particle which will be drifted due to the potential difference is electron.

We are asked to calculate the kinetic energy of the electron.

The charge of electron q = e = $-1.602\times 10^-19\ C$

The kinetic energy gained by a charged particle of charge q under a potential difference V is calculated as -

                            Kinetic energy K.E = q ×V

Hence, kinetic energy gained by the electron is -

                                     K.E = q ×V

                                            = $1.602\times 10^-19\ C\ \times 50V$

                                            = $80.1\times 10^-19\ Joule$   [ANS]

Answer 2

Answer:

The kinetic energy gained by the electron is 8 × $10^{-18} J$ .

Concept:

Kinetic Energy of an electron accelerated due to the potential difference.

Given:

Potential difference between the electrodes, V = 50 V

The particle which is accelerated due to this potential difference is an electron. It starts moving from cathode to anode. During this motion, it gains some kinetic energy.

Charge on an electron, e = $1.6 X 10^{-19} C$

Find:

Kinetic energy gained by the electron.

Solution:

Kinetic energy gained by an electron which is drifted due to potential difference is given by,

Kinetic Energy (K.E.) = Charge on electron x Potential Difference

K.E. = e x V

      =  $1.6 X 10^{-19} C$ x 50

      = $80 X 10^{-19} J$

      = 8 × $10^{-18} J$

Hence, the kinetic energy gained by the electron is 8 × $10^{-18} J$ .

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