An electron of charge 1.6 x 1019C is accelerated in vacuum from rest at zero volt towards a plate of 40KV. Calculate the kinetic energy of the electron.
4 x 10 25J
4 x 1021 J
6.4 x 1020 J
6.4 x 10-15 J
2.5 x 1020 J
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Discussions (6)
To solve this problem, we need to use the relation between the kinetic energy of a charged particle and the potential difference (voltage) through which it is accelerated.
Given information:
- Charge of the electron = 1.6 × 10^-19 C
- Accelerating potential difference = 40 kV = 40,000 V
The kinetic energy (K.E.) of a charged particle accelerated through a potential difference (V) is given by the equation:
K.E. = q × V
Where:
- q is the charge of the particle
- V is the potential difference
Substituting the given values, we get:
K.E. = (1.6 × 10^-19 C) × (40,000 V)
K.E. = 6.4 × 10^-15 J
Therefore, the kinetic energy of the electron is 6.4 × 10^-15 J.
The correct answer is D. 6.4 × 10^-15 J.
