Question

If the kinetic energy of a proton is increased by 16 times, the wavelength of the de-Broglie wave associated with it would become:

• A. 4 times
• B. $\frac{1}{16}$ times
• C. $\frac{1}{4}$ times
• D. 16 times

Answer 1

The correct option is C $\frac{1}{4}$ times

We know,
$\lambda =\frac{h}{\sqrt{2K.E\times m}}$
where 'm' is the mass of the particle
Let ${\lambda }_1$ and ${\lambda }_2$ be the initial and final wavelengths.
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{h}{\sqrt{2K.E_1\times m}}}{\frac{h}{\sqrt{2K.E_2\times m}}}$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{K{E}_2}}{\sqrt{K{E}_1}}$
But $K{E}_2=16K{E}_1$

$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{16K{E}_1}}{\sqrt{K{E}_1}}$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=4$
$\therefore {\lambda }_2=\frac{1}{4}\times {\lambda }_1$