If a free electron moves through a potential difference of 1 kV, then

If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by

[amp_mcq option1=”1.6 × 10⁻¹⁹ J” option2=”1.6 × 10⁻¹⁶ J” option3=”1 × 10⁻¹⁹ J” option4=”1 × 10⁻¹⁶ J” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2018

The energy gained or lost by a charge (q) moving through a potential difference (V) is given by ΔE = qV. The charge of a free electron is e = 1.602 × 10⁻¹⁹ Coulombs. The potential difference is 1 kV = 1000 Volts. Therefore, the energy gained is ΔE = (1.602 × 10⁻¹⁹ C) × (1000 V) = 1.602 × 10⁻¹⁶ Joules. Option B is the closest value.

This question tests the concept of electric potential energy and its relation to potential difference and charge (ΔE = qΔV). It also requires knowing the charge of an electron and unit conversions (kV to V).

The energy gained by an electron moving through a potential difference of 1 Volt is defined as 1 electronvolt (eV). So, moving through 1 kV (1000 V) gives an energy gain of 1000 eV. Since 1 eV = 1.602 × 10⁻¹⁹ J, 1000 eV = 1000 × 1.602 × 10⁻¹⁹ J = 1.602 × 10⁻¹⁶ J.

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