Question

What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?

• A. 12.27 $\mathring{A}$
• B. 1.227 $\mathring{A}$
• C. 0.1227 $\mathring{A}$
• D. 0.001227 $\mathring{A}$

Answer 1

Answer: (B)

Given : $V=100$ volts

de-Broglie wavelength of electron $\lambda =\frac{h}{\sqrt{2meV}}$

$\therefore$ $\lambda =\frac{6.6\times {10}^{-34}}{\sqrt{2(9.1\times {10}^{-31})(1.6\times {10}^{-19})\left(100\right)}}$

Or $\lambda =\frac{6.6\times {10}^{-34}}{5.4\times {10}^{-10}}$ m

$⟹$ $\lambda =1.227\times {10}^{-10}m=1.227$ $\mathring{A}$