An electron of mass 9.0 \times 10^{-31}\text{ kg} and charge 1.6 \times 10^{-19}\text{ C} … — JAMB UTME

An electron of mass $9.0 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$ is accelerated from rest through an electric potential difference $V$ . If the associated de Broglie wavelength of the electron is $1.1 \times 10^{-10}\text{ m}$ and Planck's constant is $6.6 \times 10^{-34}\text{ J s}$ , what is the value of the potential difference $V$ ?

$125\text{ V}$ Answer
B
$250\text{ V}$
C
$11.2\text{ V}$
D
$62.5\text{ V}$

Answer

The accelerating potential difference required is

125\text{ V}

The de Broglie wavelength of an electron accelerated from rest through a potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2meV}}

. Rearranging for

V

gives

V = \frac{h^2}{2me\lambda^2}

. Substituting

h = 6.6 \times 10^{-34}\text{ J s}

m = 9.0 \times 10^{-31}\text{ kg}

e = 1.6 \times 10^{-19}\text{ C}

, and

\lambda = 1.1 \times 10^{-10}\text{ m}

gives

V = 125\text{ V}

Step-by-Step Solution

Relate kinetic energy to momentum and accelerating potential.

E_k = eV

and

p = \sqrt{2m E_k} = \sqrt{2meV}

An electron accelerated through potential

V

gains kinetic energy equal to

eV

Substitute momentum into the de Broglie wavelength equation.

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}

De Broglie wavelength is defined as Planck's constant divided by momentum.

Rearrange the equation to solve for potential difference

V

V = \frac{h^2}{2me\lambda^2}

Squaring both sides allows isolating

V

Substitute the given numerical values and calculate

V

$V = \frac{(6.6 \times 10^{-34})^2}{2(9.0 \times 10^{-31})(1.6 \times 10^{-19})(1.1 \times 10^{-10})^2} = 125\text{ V}$

Performing the algebraic substitution yields the final potential.

Key Concept

de Broglie wavelength of an electron accelerated through a potential difference

Estimated Time:1m 30s

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