Question

Calculator the de-Broglie wavelength of an electron beam acceleration through a potential difference of $60V$.

• A. $1.58A$
• B. $2.58A$
• C. $3.58A$
• D. $4.58A$

Answer 1

The correct option is A $1.58A$

Potential difference $\Delta V=60$ so energy $E=q\Delta V=1.6\times {10}^{-19}\times 60Joule=9.6\times {10}^{-18}Joule$

so de-Broglie wavelength $\lambda =\frac{h}{\sqrt[2]{22mE}}$ .......... (1)

as momentum, $p=mv=\sqrt[2]{22mE}$

putting $m=9.1\times {10}^{-31}Kg$ and $h=6.62\times {10}^{-34}Js$ in eqaution-1

we get , $\lambda =1.58\times {10}^{-10}meter=1.58Angstrom$