Question

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.)

Short answer

De Broglie wavelength is $7.75\text{pm}$.

Step-by-step solution

A concept:

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.

Formula used:

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

The known data:

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" $e=1.602\times {10}^{-19}\text{\hspace{0.17em}C}$

Wavelength:

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 $\begin{array}{rcl}& & 7.75\text{pm}\end{array}$.