91Ó°ÊÓ

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.

$\mathrm{λ}=\frac{h}{p}$

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

Hence, the wavelength is,

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

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" $h=6.626×{10}^{−34}\text{ J}â‹\dots \text{s}$

Mass of electron, localid="1664289984939" $m_e=9.109×{10}^{−31}\text{ k²µ}$

Charge of electron,localid="1664289989103" $e=1.602×{10}^{−19}\text{ C}$

Define the de Broglie wavelength as below.

$\begin{array}{rcl}\mathrm{λ}& =& \frac{h}{\sqrt{2m_{e}eV}}\\ & =& \frac{6.626×{10}^{−34}\text{ J}â‹\dots \text{s}}{\sqrt{2(9.109×{10}^{−31}\text{ k²µ})(1.602×{10}^{−19}\text{ C})(25×{10}^3\text{ V})}}\\ & =& 7.75×{10}^{−12}\text{″¾}\\ & =& 7.75\text{pm}\end{array}$

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