Question

The wavelength of the yellow spectral emission line of sodium is $\mathbf{\text{590nm}}$. At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?

Short answer

The kinetic energy would be $4.32\times {10}^{-6}\text{\hspace{0.17em}eV}$.

Step-by-step solution

The given data:

The wavelength of the yellow spectral emission line of sodium is$\text{590nm}$.

Definition and Concept of de Broglie wavelength:

The wavelength that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength

The greater the potential difference, the greater the kinetic energy and momentum, and smaller is the de Broglie wavelength.

The de Broglie wavelength is defined by,

$\lambda =\frac{h}{p}$ ….. (1)

Here, $p$is momentum of particle and $h$is Plank’s constant having a value $6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$.

Write the equation for the kinetic energy of the electron as,

$K=\frac{p^2}{2m}$ ….. (2)

Here, $p$is electron momentum and $m$is mass of electron having a value $9.109\times {10}^{-31}\text{\hspace{0.17em}kg}$.

Thus, electron momentum will be,

$p=\sqrt{2mK}$ ..... (3)

Determining the de Broglie wavelength λ:

As de Broglie wavelength is,

$\lambda =\frac{h}{p}$

Substitute equation (3) in the above equation.

$\begin{array}{rcl}\lambda & =& \frac{h}{\sqrt{2mK}}\\ & =& \frac{6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{\sqrt{2(9.109\times {10}^{-31}\text{\hspace{0.17em}kg})\left(1.602\times {10}^{-19}{\displaystyle \raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{eV}$}\right.}\right)K}}\\ & =& \frac{1.226\times {10}^{-9}\text{\hspace{0.17em}m}\cdot {\text{eV}}^{1/2}}{\sqrt{K}}\end{array}$

$\lambda =\frac{1.226\text{\hspace{0.17em}nm}\cdot {\text{eV}}^{1/2}}{\sqrt{K}}$ ..... (4)

Determining the kinetic energy:

Now, with wavelength, $\lambda =\text{590\hspace{0.17em}nm}$

Kinetic energy can found by substituting the above value of wavelength into equation (4).

role="math" localid="1663137322393" $\begin{array}{rcl}K& =& {\left(\frac{1.226\text{\hspace{0.17em}nm}\cdot {\text{eV}}^{1/2}}{\lambda }\right)}^2\\ & & {=\left(\frac{1.226\text{\hspace{0.17em}nm}\cdot {\text{eV}}^{1/2}}{590\text{\hspace{0.17em}nm}}\right)}^2\\ & =& 4.32\times {10}^{-6}\text{\hspace{0.17em}eV}\end{array}$

Hence, the kinetic energy is $\begin{array}{rcl}& & 4.32\times {10}^{-6}\text{\hspace{0.17em}eV}\end{array}$.