Question

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

• A. a decrease of 11%
• B. an increase of 11.1%
• C. an increase of 10%
• D. an increase of 23.4%

Answer 1

The correct option is D

an increase of 23.4%

Using the relation, $\lambda \propto \frac{1}{K.E.},$ we get
${\lambda }_1=\lambda$ and $K.E{.}_1=E$ (initial)
${\lambda }_2=0.9\lambda$
[10% decrease from ${\lambda }_1,{\lambda }_2=\lambda -\frac{10}{100}\lambda =\lambda (1-0.1)=0.9\lambda ]$
We are required to find $K.E_2$ = ?
$\therefore K.E{.}_2=\frac{{\lambda }_1^2}{{\lambda }_2^2}\times K.E{.}_1=\frac{{\lambda }^2}{(.9\lambda {)}^2}\times K.E{.}_1$
$=\frac{{\lambda }^2\times 100}{81{\lambda }^2}\times E$ $[K.E_1=E]$
$=\frac{100}{81}E$ [more than $K.E{.}_1$ i.e., increase]
Now increase in $K.E{.}_2-K.E{.}_1=\frac{100}{81}E-E$
Or % increase of $K.E.=\frac{\frac{100}{81}E-E}{E}\times 100$
$=\frac{19}{81}\times 100=23.4$