In a cathode-ray tube, electrons of mass 9.10 \times 10^{-31}\text{ kg} and elementary cha… — JAMB UTME

In a cathode-ray tube, electrons of mass $9.10 \times 10^{-31}\text{ kg}$ and elementary charge $1.60 \times 10^{-19}\text{ C}$ are accelerated from rest by the electric field between the cathode and the anode. What potential difference, in volts, is required to accelerate these electrons to a speed of $8.00 \times 10^{6}\text{ m s}^{-1}$ ?

Answer: 182 V

Answer

The potential difference required to accelerate the electrons to the specified speed is

182\text{ V}

The work done by an accelerating potential difference

V

on an electron of charge

e

is converted entirely into kinetic energy

\frac{1}{2} m v^2

. Solving

e V = \frac{1}{2} m v^2

for

V

yields

V = \frac{m v^2}{2 e} = 182\text{ V}

Step-by-Step Solution

Equate the work done by the electric field to the kinetic energy gained by an electron.

W = e V = \frac{1}{2} m v^2

Work done on a charged particle moving through an electric potential difference equals its gain in kinetic energy.

Isolate the potential difference

V

on one side of the equation.

V = \frac{m v^2}{2 e}

Algebraic rearrangement to solve for the target variable.

Substitute the given numerical parameters into the equation.

V = \frac{(9.10 \times 10^{-31}\text{ kg}) \times (8.00 \times 10^{6}\text{ m s}^{-1})^2}{2 \times (1.60 \times 10^{-19}\text{ C})}

Populating the formula with the specified values for electron mass, speed, and charge.

Perform the final calculation.

V = 182\text{ V}

Simplifying the numerical expression gives

182\text{ V}

Key Concept

Acceleration of charged particles in electric fields

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