91影视

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

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,

$位=\frac{h}{p}$ 鈥�.. (1)

Here, $p$is momentum of particle and $h$is Plank鈥檚 constant having a value $6.626脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s}$.

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

$K=\frac{p^2}{2m}$ 鈥�.. (2)

Here, $p$is electron momentum and $m$is mass of electron having a value $9.109脳{10}^{鈭�31}\text{鈥塳驳}$.

Thus, electron momentum will be,

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

As de Broglie wavelength is,

$位=\frac{h}{p}$

Substitute equation (3) in the above equation.

$\begin{array}{rcl}位& =& \frac{h}{\sqrt{2mK}}\\ & =& \frac{6.626脳{10}^{鈭�34}\text{鈥塉}鈰�\text{s}}{\sqrt{2(9.109脳{10}^{鈭�31}\text{鈥塳驳})\left(1.602脳{10}^{鈭�19}{\displaystyle \raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{eV}$}\right.}\right)K}}\\ & =& \frac{1.226脳{10}^{鈭�9}\text{鈥尘}鈰�{\text{eV}}^{1/2}}{\sqrt{K}}\end{array}$

$位=\frac{1.226\text{鈥塶尘}鈰�{\text{eV}}^{1/2}}{\sqrt{K}}$ ..... (4)

Now, with wavelength, $位=\text{590鈥塶尘}$

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

role="math" localid="1663137322393" $\begin{array}{rcl}K& =& {\left(\frac{1.226\text{鈥塶尘}鈰�{\text{eV}}^{1/2}}{位}\right)}^2\\ & & {=\left(\frac{1.226\text{鈥塶尘}鈰�{\text{eV}}^{1/2}}{590\text{鈥塶尘}}\right)}^2\\ & =& 4.32脳{10}^{鈭�6}\text{鈥塭痴}\end{array}$

Hence, the kinetic energy is $\begin{array}{rcl}& & 4.32脳{10}^{鈭�6}\text{鈥塭痴}\end{array}$.