Question

Find the:
i) momentum
ii) speed
iii) de Broglie wavelength
of an electron with the kinetic energy of 120 eV.
The given values are, $h=6.63\times {{10}^{-34}}Js$, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$, and $1eV=1.6\times {{10}^{-19}}J$

Answer 1

Recall the formulae that are required to calculate all, the momentum, velocity, and the wavelength. The momentum can be calculated only after you find the velocity and the wavelength can be calculated only after finding the momentum.

Complete step by step solution:
We have been provided with the kinetic energy of the electron. A formula that incorporates the kinetic energy as well as the velocity is:
$K.E.=\dfrac{1}{2}{{m}_{e}}{{v}^{2}}$
Where, $v$ is the velocity or speed of the electron, ${{m}_{e}}$ is the given mass, and $K.E.$ is the given kinetic energy, we will convert it into joules while calculating the velocity. Now, putting the values in the equation and solving for velocity, we get:

& {{v}^{2}}=\dfrac{K.E.\times 2}{{{m}_{e}}} \\\ & {{v}^{2}}=\dfrac{2\times 120\times 1.6\times {{10}^{-19}}J}{9.1\times {{10}^{-31}}kg} \\\ & {{v}^{2}}=42.196\times {{10}^{12}}{{m}^{2}}/{{s}^{2}} \\\ & v=\sqrt{42.196\times {{10}^{12}}}m/s \\\ & v=6.495\times {{10}^{6}}m/s \\\ \end{aligned}$$ Now that we have the velocity of the electron, we can easily find its momentum, since momentum is the product of mass and velocity. The formula is: $$p={{m}_{e}}v$$ Where, $p$ denotes the momentum and the other variables denote meanings described previously. So, solving for momentum, we get: $$\begin{aligned} & p=(9.1\times {{10}^{-31}}kg)\times (6.495\times {{10}^{6}}m/s) \\\ & p=5.910\times {{10}^{-24}}kgm{{s}^{-1}} \\\ \end{aligned}$$ The de Broglie wavelength is defined as the Planck’s constant divided by the momentum. The formula is as follows: $$\lambda =\dfrac{h}{p}$$ Where, $\lambda $is the de Broglie wavelength, $h$ is the Planck’s constant. Now, solving for lambda, we get: $$\begin{aligned} & \lambda =\dfrac{6.63\times {{10}^{-34}}Js}{5.910\times {{10}^{-24}}kgm{{s}^{-1}}} \\\ & \lambda =1.122\times {{10}^{-10}}m \\\ \end{aligned}$$ Thus, we have all the required values. **Note:** It is necessary to convert the electron volts into joules since all the other parameters are present in their SI units and calculations become easier in SI units. We can convert the final wavelength to angstroms by multiplying with ${{10}^{10}}$. So, the wavelength of the electron is $1.122{{A}^{0}}$.