OCR.4.1.3

electron volt (eV)

Energy, Power & Resistance - OCR A-Level Physics

Electrical Energy

$$W = IVt$$

EnergyThe capacity to do work. Measured in joules (J). can also be measured in kilowatt-hoursA unit of energy equal to 3.6 × 10⁶ J, commonly used for electrical energy billing. (kWh).
1 $kWh = energyThe capacity to do work. Measured in joules (J).$ used by a 1 kW appliance running for 1 hour.
1 $kWh = 1000 W x 3600 s = 3.6 MJ$.
$Cost = energyThe capacity to do work. Measured in joules (J). (kWh) \times price per kWh.$

Accelerating Charged Particles

A PD between two points accelerates charged particles.
Energy gained by a chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).: $E = eV (chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). x PD)$.
If starting from rest, all energy becomes kinetic energyThe energy an object possesses due to its motion..

eV = \frac{1}{2}mv^{2}

Key Definition

electron volt (eV)
The energy gained by an electron accelerated through a PD of 1 V. 1 eV = $1.6 \times 10^{-19}$ J.

An electron gunA device that accelerates electrons through a potential difference, giving them kinetic energy eV. accelerates electrons using a PD.
The electron voltThe SI unit of potential difference and EMFElectromotive force. The energy transferred per unit charge by a source in driving charge around a complete circuit. Measured in volts (V).. One volt is one joule per coulomb.The energy gained by an electron accelerated through a potential differenceThe energy transferred per unit chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). between two points. Measured in volts (V). of 1 V. Equal to 1.6 × 10⁻¹⁹ J. is a convenient unit for small energies at atomic scale.
To convert eV to J: multiply by $1.6 \times 10^{-19}$.
To convert J to eV: divide by $1.6 \times 10^{-19}$.

Worked Example

An electron is accelerated from rest through a PD of 200 V. Calculate its final speed. (m_e = $9.11 \times 10^{-31}$ kg, e = $1.6 \times 10^{-19}$ C)

Show Solution

Energy gained

$E = eV = 1.6 x 10^{-19} x 200 = 3.2 x 10^{-17} J$

Set equal to KE

$3.2 x 10^{-17} = 1/2 x 9.11 x 10^{-31} x v^2$

Rearrange

$v^2 = (2 x 3.2 x 10^{-17}) / (9.11 x 10^{-31}) = 7.03 x 10^{13}$

$v = sqrt(7.03 x 10^{13}) = 8.4 x 10^{6} m/s$

Answer

$v = 8.4 x 10^{6} $m $s^{-1}$$ (2 s.f.)$

Common Mistake MEDIUM

Using the wrong charge in $eV = 1/2 mv^2. Use$ the charge of the particle being accelerated (e.g. $1.6 \times 10^{-19}$ C for a single electron or proton), not 1 eV.

Related:Waves Photon Energy Particle Physics