Question

Singly charged sodium ions are accelerated through a potential difference of $\mathbf{\text{300\hspace{0.17em}V}}$. (a) What is the momentum acquired by such an ion? (b) What is its de Broglie wavelength?

Short answer

(a) The momentum is $1.91\times {10}^{-21}\raisebox{1ex}{${\displaystyle \text{kg}\cdot \text{m}}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$.

(b) De Broglie wavelength is $3.46\times {10}^{-13}\text{\hspace{0.17em}m}$.

Step-by-step solution

The given data:

Singly charged sodium ions are accelerated through a potential difference of$300\mathrm{V}$.

Definition 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 de Broglie wavelength is defined by using following formula.

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

Here, $h$is Planck’s constant and $p$is the momentum.

Kinetic energy of ion is,

$K=qV$

Here, $q$ is the charge having a value $1.6\times {10}^{-19}\text{\hspace{0.17em}C}$and $V$is the potential having a value $\text{300\hspace{0.17em}V}$.

Write the equation for the momentum as,

$p=\sqrt{2mK}$

Here, $m$ is mass of ion and $K$ is Kinetic energy.

(a) Determining the momentum of ion:

The kinetic energy of the ion is,

$\begin{array}{rcl}K& =& qV\\ & =& (1.6\times {10}^{-19}\text{\hspace{0.17em}C})\left(300\text{\hspace{0.17em}V}\right)\\ & =& 4.8\times {10}^{-17}\text{\hspace{0.17em}J}\end{array}$

As the mass of single sodium ion is $3.819\times {10}^{-26}\text{\hspace{0.17em}kg}$.

Thus, momentum of a sodium ion is,

role="math" localid="1663135253292" $\begin{array}{rcl}p& =& \sqrt{2mK}\\ & =& \sqrt{2(3.819\times {10}^{-26}\text{\hspace{0.17em}kg})(4.8\times {10}^{-17}\text{\hspace{0.17em}J})}\\ & =& 1.91\times {10}^{-21}\raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$

Hence, the momentum of ion is $\begin{array}{rcl}& & 1.91\times {10}^{-21}\raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\end{array}$.

(b) Determining the de Broglie wavelength:

The de Broglie wavelength is,

$\begin{array}{rcl}\lambda & =& \frac{h}{p}\\ & =& \frac{6.63\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{1.91\times {10}^{-21}{\displaystyle \raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}\\ & =& 3.46\times {10}^{-13}\text{\hspace{0.17em}m}\end{array}$

Hence, de Broglie wavelength is $\begin{array}{rcl}& & 3.46\times {10}^{-13}\text{\hspace{0.17em}m}\end{array}$.