Question

Electrons used in an electron microscope are accelerated by a voltage of $25$ kV. If the voltage is increased to $100$ kV then the de-Broglie wavelength associated with the electrons would?

• A. Increase by $2$ times
• B. Decrease by $2$ times
• C. Decrease by $4$ times
• D. Increase by $4$ times

Answer 1

The correct option is B Decrease by $2$ times

De Broglie wavelength in terms of accelerated potential difference is

$\lambda =\frac{12.27}{\sqrt{V}}$

Initial and final wavelength are ${\lambda }_1$and ${\lambda }_2$having potential as $V_1$and $V_2$

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{V_2}{V_1}}$

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{100}{25}}$

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{2}$

$\frac{{\lambda }_1}{{\lambda }_2}=2$

${\lambda }_2=\frac{{\lambda }_1}{2}$

So, new wavelength is decrease by 2.