In a cathode-ray experiment, electrons are accelerated from rest through a potential diffe… — JAMB UTME

In a cathode-ray experiment, electrons are accelerated from rest through a potential difference of $100\text{ V}$ . Taking the specific charge ( $\frac{e}{m}$ ) of an electron to be $1.80 \times 10^{11}\text{ C/kg}$ , calculate the final speed of the electrons as they pass through the anode aperture, expressed in units of $10^6\text{ m/s}$ .

Answer: 6 10^6 m/s

Answer

The final speed of the electrons is

6.0 \times 10^6\text{ m/s}

, which corresponds to a coefficient value of 6.0.

By applying conservation of energy, the kinetic energy acquired by electrons in a cathode-ray tube equals the electric work done on them (

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

). Solving for velocity gives

v = \sqrt{2V(e/m)}

. Substituting

V = 100\text{ V}

and

e/m = 1.80 \times 10^{11}\text{ C/kg}

gives

v = 6.0 \times 10^6\text{ m/s}

Step-by-Step Solution

Set up the energy conservation relation for cathode ray electrons.

Electrical work done

W = eV

equals kinetic energy

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

Electrons starting from rest gain kinetic energy equal to the electrical potential energy lost across the potential difference.

Rearrange the formula to isolate the electron velocity

v

v^2 = 2V\left(\frac{e}{m}\right) \implies v = \sqrt{2V\left(\frac{e}{m}\right)}

Isolating velocity allows direct calculation using the given specific charge

(\frac{e}{m})

and accelerating voltage

V

Substitute given values into the equation and compute

v

v = \sqrt{2 \times 100 \times 1.80 \times 10^{11}} = \sqrt{36 \times 10^{12}} = 6.0 \times 10^6\text{ m/s}

Evaluating the square root yields the speed in meters per second.

Key Concept

Electron acceleration in cathode ray tubes and specific charge relation

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