Question

An electron, practically at rest, is initially accelerated through a potential difference of $100$ volts. It then has a de Broglie wavelength $={\lambda }_{1}A$. It then get retarded through $19$ volts and then has a wavelength ${\lambda }_2$. A further retardation through $32$ volts changes the wavelength to ${\lambda }_{3}A$, what is $\frac{{\lambda }_3-{\lambda }_1}{{\lambda }_1}$?

Answer 1

According to de-broglie equation;

Wavelength associated with electron,

$\lambda =\frac{12.27}{\sqrt{V}}{A}^o=\frac{12.27}{\sqrt{100}}{A}^o=\frac{12.27}{10}=1.227A^o$

and the wavelength corresponds to X-rays.