Test Index

WAEC Physics 2020 Paper

© africaexams.com

Question : 41

Total: 51

An electron of mass

9.1 \times 10^{31} \mathrm{kg}

is travelling at a speed of

2.0 \times 10^{6} \mathrm{m/s}

. Calculate the associate wavelength of the electron. [h

=6.6 . \times 10^{34} \mathrm{Js} ]

$3.63 \times 10^{10} \mathrm{m}$
$3.63 \times 10^{-8} \mathrm{m}$
$3.63 \times 10^{-7} \mathrm{m}$
$6.89 \times 10^{-4} \mathrm{m}$

Solution:

Concept:

Use the de Broglie wavelength relation:

\lambda = \frac{h}{mv}

.

Explanation:

Given:

m = 9.1 \times 10^{-31}\ \mathrm{kg}

,

v = 2.0 \times 10^{6}\ \mathrm{m/s}

,

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

.

Substitute into

\lambda = \frac{h}{mv}

.

\lambda = \frac{6.6 \times 10^{-34}}{(9.1 \times 10^{-31})(2.0 \times 10^{6})}

\lambda = \frac{6.6 \times 10^{-34}}{1.82 \times 10^{-24}}

\lambda = 3.63 \times 10^{-10}\ \mathrm{m}

Answer:

The associated wavelength of the electron is

3.63 \times 10^{-10}\ \mathrm{m}

.

© africaexams.com

Go to Question:

Prev Question

Next Question