In an old-fashioned television set, electrons are accelerated through a potential difference of $\mathbf{\text{25.0kV}}$. What is the de Broglie wavelength of such electrons? (Relativity is not needed.)

The wavelength that is associated with an object in relation to its momentum and mass is known as the de Broglie wavelength. The de Broglie wavelength of a particle is usually inversely proportional to its strength.

Momentum of electron is,

$p=\sqrt{2m_{e}eV}$

Where, $V$is accelerating potential, $e$is the fundamental charge, $m_e$is mass of electron.

Wavelength is defined by using following formula.

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

Where, $h$ is plank’s constant and is momentum of electron.

Hence, the wavelength is,

localid="1664289974674" $\begin{array}{rcl}\lambda & =& \frac{h}{p}\\ & =& \frac{h}{\sqrt{2m_{e}eV}}\end{array}$

In an old-fashioned television set, electrons are accelerated through a potential difference of $\text{25.0kV}$. Therefore,

$V=\text{25.0kV}$

Plank’s constant, localid="1664289980563" $h=6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$

Mass of electron, localid="1664289984939" $m_e=9.109\times {10}^{-31}\text{\hspace{0.17em}kg}$

Charge of electron,localid="1664289989103" $e=1.602\times {10}^{-19}\text{\hspace{0.17em}C}$

Define the de Broglie wavelength as below.

$\begin{array}{rcl}\lambda & =& \frac{h}{\sqrt{2m_{e}eV}}\\ & =& \frac{6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}}{\sqrt{2(9.109\times {10}^{-31}\text{\hspace{0.17em}kg})(1.602\times {10}^{-19}\text{\hspace{0.17em}C})(25\times {10}^3\text{\hspace{0.17em}V})}}\\ & =& 7.75\times {10}^{-12}\text{\hspace{0.17em}m}\\ & =& 7.75\text{pm}\end{array}$

Hence, required de Broglie wavelength is $\begin{array}{rcl}& & 7.75\text{pm}\end{array}$.