Question

An electron is continuously accelerated in a vacuum tube by applying potential difference. If the de Broglie's wavelength is decreased by $10\%$, the change in the kinetic energy of the electron is (nearly):

• A. decreased by $11\%$
• B. increased by $23.4\%$
• C. increased by $10\%$
• D. increased by $11.1\%$

Answer 1

The correct option is B increased by $23.4\%$

We know that, $\lambda =\frac{h}{\sqrt{2mkE}}$

$\Rightarrow \lambda \propto \frac{1}{\sqrt{kE}}$.........(1)

Let ${\lambda }_1=\lambda ,K{E}_1=E$.........(1)

$\therefore$ according to question, ${\lambda }_2=0.9\lambda$..........(2)

( $\lambda$ is decreased by $10\%$)

$K{E}_2=?$

$\Rightarrow K{E}_2=\frac{{\lambda }_1^2}{{\lambda }_2^2}\times K{E}_1=\frac{{\lambda }^2}{(0.9\lambda {)}^2}\times E$

$=\frac{100E}{81}$ [Putting values from eq.(1) and (2)]

$\therefore \%$ increase $=\frac{K{E}_2-K{E}_1}{K{E}_1}\times 100$

$=\frac{\frac{100}{81}E-E}{E}\times 100$

$=\frac{19}{81}\times 100$

$=23.4\%$

$\therefore$ The kinetic energy increases by $23.4\%$