Question

If the kinetic energy of an electron is increased four times, then what would be the wavelength of the de-Broglie wave associated with it.

Answer 1

K.E$=\frac{p^2}{2m}$

K.E$\alpha p^2$

$\because$ K.E is increased four times

$\therefore$ P is increased two times

$\lambda =\frac{h}{p}$

$\lambda \alpha \frac{1}{p}$

Wavelength is decreased two times.