Question

the de Broglie wavelength associated with an electron accelerated through a potential difference of 100V is

Answer 1

Let the charge of electron be α (α =1.6x 10^-19 )

potential difference = Workdone/α

workdone = potential difference X α

work done in this case is equal to kinetic energy which is 1/2 Mv²

where M is mass of electron and V is velocity of electron

1/2 mv² = 100 x α

multiplying and dividing LHS by M

1/2 m²v²/2m =100α

we know m²v² is p² where p is momentum

1/2 p²/2m =100 α

(m= 9.1x10^-31)

form there find value of p and put in eq...

lamda = h/p

lamda is wavelength
h is planks constant

if u dont get anything ask me in comment

Answer 2

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$

• Acceleration potential ${V=100 v}$ .The de Broglie wavelength $\lambda{is}$
• $\lambda{h/p}$=$\frac{1.227}{\sqrt{V}}$ nm
• $\lambda$=$\frac{1.227}{\sqrt{100}}$ nm=${0.124nm}$.
• The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.