Question

Find the de Broglie wavelength of an electron having energy $10{\text{keV}}$ .
A) $12{\text{A}}^\circ$
B) $1 \cdot 2{\text{A}}^\circ$
C) $12 \cdot 2{\text{A}}^\circ$
D) None of these

Answer 1

de Broglie came up with the concept that matter exhibits dual nature i.e wave nature and particle nature. the de Broglie wavelength of the electron is the wavelength associated with the electron having a mass and momentum. The energy of this electron will be inversely proportional to the de Broglie wavelength of the electron.

Formula used:
-The energy of a quantum particle is given by, $E = \dfrac{{hc}}{\lambda }$ where $h$ is Planck's constant, $c$ is the speed of light and $\lambda$ is de Broglie wavelength of the particle.

Step by step solution .
Step 1: Express the relation for the energy of the given electron.
An electron is a subatomic particle and can be considered as a quantum particle.
Let $\lambda$ be the de Broglie wavelength of the electron whose energy is given to be $E = 10{\text{keV}} = {10^4}{\text{eV}}$.
Then the energy of the given electron can be expressed by the relation $E = \dfrac{{hc}}{\lambda }$ where $h$ is Planck's constant and $c$ is the speed of light. This relation suggests that the energy of the electron is inversely proportional to its de Broglie wavelength.
From the above relation, we have the de Broglie wavelength of the electron as $\lambda = \dfrac{{hc}}{E}$ ------ (1)
Step 2: Make appropriate substitutions in equation (1) to obtain the value of the de Broglie wavelength of the given electron.
Equation (1) is given as $\lambda = \dfrac{{hc}}{E}$ .
Substituting for $h = 6 \cdot 626 \times {10^{ - 34}}{\text{Js}}$ , $c = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ and $E = {10^4} \times 1 \cdot 6 \times {10^{ - 19}}{\text{J}}$ in equation (1) we get, $\lambda = \dfrac{{6 \cdot 626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{{{10}^4} \times 1 \cdot 6 \times {{10}^{ - 19}}{\text{J}}}} = 1 \cdot 2 \times {10^{ - 10}}{\text{m}}$
Thus we obtain the de Broglie wavelength of the given electron to be $\lambda = 1 \cdot 2{\text{A}}^\circ$ .

Hence the correct option is B.

Note: While substituting values in any equation, make sure that all the quantities are expressed in their respective S.I units. If not, then the necessary conversion of the units must be done. Here the energy of the electron was expressed in kilo electron-volts or electron-volts. So we convert this quantity into the unit of joules as $E = {10^4} \times 1 \cdot 6 \times {10^{ - 19}}{\text{J}}$ since $1{\text{eV}} = 1 \cdot 6 \times {10^{ - 19}}{\text{J}}$ , before substituting in equation (1). Also ${10^{ - 10}}{\text{m}} = 1{\text{A}}^\circ$ so we have the de Broglie wavelength as $\lambda = 1 \cdot 2 \times {10^{ - 10}}{\text{m}} = 1 \cdot 2{\text{A}}^\circ$ .