If we apply potential difference so that an electron is accelerated continuously in a vacuum tube such that a decrease of 10% occurs in its de-Broglie wave length. In such a case the change observed in kinetic energy of electron will be approximately

Using the relation, $\lambda \propto \frac{1}{\text{ K.E. }}\text{, we get }$

$\begin{array}{l}{\lambda }_1=\lambda \text{ and }\mathrm{K}.\mathrm{E}.1=E\text{ (initial) }\\ {\lambda }_2=0.9\lambda \end{array}$

$\begin{array}{l}\left[10\mathrm{\%}\text{ decrease from }{\lambda }_1,{\lambda }_2=\lambda -\frac{10}{100}\lambda \right.\\ =\lambda (1-0.1)=0.9\lambda ]\end{array}$

We are required to find K.E.₂ =?

$\begin{array}{l}\therefore \mathrm{K}.{\mathrm{E}}_{\cdot 2}=\frac{{\lambda }_1^2}{{\lambda }_2^2}\times \mathrm{K}.{\mathrm{E}}_{\cdot }\cdot \frac{{\lambda }^2}{(.9\lambda {)}^2}\times \mathrm{K}.{\mathrm{E}}_1\\ =\frac{{\lambda }^2\times 100}{81{\lambda }^2}\times E\left[\mathrm{K}.{\mathrm{E}}_1=E\right]\\ =\frac{100}{81}E\left[\text{ more than }\mathrm{K}.{\mathrm{E}}_{.1}\text{ i.e., increase }\right]\end{array}$

Now increase in $\mathrm{K}.\mathrm{E}.=\mathrm{K}.{\mathrm{E}}_{.2}-\mathrm{K}.{\mathrm{E}}_{.1}=\frac{100}{81}E-E$

$\mathrm{\%}\text{ increase of }\mathrm{K}.\mathrm{E}.=\frac{\frac{100}{81}E-E}{E}\times 100$

$=\frac{19}{81}\times 100=23.4\mathrm{\%}$