Question

What is the speed and De Broglie wave length of electron has been accelerated by a potential difference 500V.​

Answer 1

$\underline{\underline{\Huge{\textsf{Concept Behind : }}}}$

$\hspace{1cm} \maltese$If an Electron accelerated through a Potential Difference of V Volts, then  the electron is said to possess Kinetic Energy as a result of acceleration through a Potential Difference.

This Kinetic Energy is given by the Equation :

$\LARGE{\boxed{\quad\bigstar\; \; \mathbf{\dfrac{1}{2}\, mv^2 = eV}\qquad}}$

Where,

m= $\rm Mass\: of\: the\: electron\: = 9.1 \times 10^{-31}\: \rm kg$

e = $\rm Charge\: of\: electron =1.602 \times 10^{-19}\: C$

v = $\rm Velocity\: of\:the\:electron\: (m/s)$

V = $\rm Potential \: Difference\: (V)$

Rewriting Equation for Velocity, we get

$\LARGE{\boxed{\quad\bigstar\; \; \mathbf{v = \sqrt{\dfrac{2eV}{m}}}\qquad}}$


$\hspace{1cm} \maltese$De broglie wavelength of an Electron is given by

$\LARGE{\boxed{\quad\bigstar\; \; \mathbf{\lambda = \dfrac{h}{mv}}\qquad}}$

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

Given,

Electron has been accelerated by a potential difference 500V

$\underline{\LARGE{\texttt{Step-1:}}}$

$\sf\textsf{Find the Speed of the Electron}$

$\sf\textsf{Substitute Values}$

$\begin{aligned}\sf Speed\: of\: the\: Electron, v &= \sqrt{\dfrac{2\times 1.602\times10^{-19}\times500}{9.1\times10^{-31}}}\\\\\impliesv&=\sqrt{1000\times \dfrac{1.602}{9.1}\times 10^{(-19+31)}}\\\\\impliesv&=\sqrt{10^{3}\times \dfrac{1.602}{9.1}\times 10^{12}}\\\\\implies v &=\sqrt{\cancel{10^{3}}\times \dfrac{1602\times \cancel{10^{3}}}{91\times 10^{1}}\times 10^{12}}\\\\\impliesv&=\sqrt{\dfrac{1602}{91}\times 10^{1}\times 10^{12}}\end{aligned}$

$\implies \begin{aligned}v &= \sqrt{\dfrac{1602}{91}\times 10^{13}}\end{aligned}$

$\left(\dfrac{1602}{91} = 17.60\right)$

$\begin{aligned}\implies v&=\sqrt{17.60\times10^{13}}\\\\\implies v &= \sqrt{17.6 \times 10^{13}}\\\\\implies v&=\sqrt{(176\times\cancel{10^{-1}})\times(\cancel{10^{1}}\times 10^{12})}\\\\\implies v &= \sqrt{176\times 10^{12}}\\\\\implies v &= \sqrt{176}\times 10^{6}\end{aligned}$

$\left(\sqrt{176}= 13.26\right)$

$\implies v = 13.26\times 10^{6} m/s$

$\underline{\LARGE{\texttt{Step-2:}}}$

$\sf\textsf{Find the Debroglie wavelength of the Electron}$

$\sf\textsf{Substitute Values}$

$\begin{aligned}\sf Debroglie\: wavelength\; \lambda &= \dfrac{6.626\times 10^{-34}}{(9.1\times10^{-31})\times(13.26\times10^{6})}\\\\\lambda &= \dfrac{6.626\times 10^{-34}}{(9.1\times13.26\times10^{-25})}\\\\\lambda &= \dfrac{6.626}{9.1\times13.26}\times 10^{-34+25}\\\\\lambda &= \dfrac{6.626}{120.666}\times 10^{-34+25}\\\\\lambda &= \dfrac{6.626}{120.666}\times 10^{-9}\end{aligned}$

$\begin{aligned}\implies \sf Debroglie\:wavelength\;\lambda &= \dfrac{6.626}{120.6}\times 10^{-9}\\\\\implies\lambda &= \dfrac{6626 \times 10^{-3}}{1206\times 10^{-1}}\times 10^{-9}\\\\\implies\lambda &= \dfrac{6626}{1206} \times 10^{-2}\times 10^{-9}\\\\\implies\lambda &= \dfrac{6626}{1206} \times 10^{-11}\end{aligned}$

$\left(\dfrac{6626}{1206}= 5.494\right)$

$\implies \mathsf{\lambda = 5.494 \times 10^{-11}\:m}$

$\implies \mathsf{\lambda = 0.5494 \times 10^{-10}\:m}$

$\textsf{Expressing in Angstroms}$

$\LARGE{\boxed{\quad\bigstar\; \; \mathbf{1 \AA=1\times{10}^{-10}\:m}\qquad}}$

$\implies \mathsf{\lambda = 0.5494 \AA}$

$\therefore$

$\blacksquare \; \: \textsf{Speed of the Electron = }\underline{\Large{\mathbf{13.26\times 10^{6}\: m/s }}}$

$\blacksquare\, \: \textsf{Wavelength of the Electron = }\underline{\Large{\mathbf{5.494\times10^{-11}\: m=0.5494 \AA}}}$