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Q.
If the kinetic energy of an electron is increased four times, the wavelength of the de-Broglie wave associated with it would become
AIEEEAIEEE 2012Structure of Atom
Solution:
de - Broglie wavelength is given by :
$\lambda=\frac{h}{mv}\,...\left(i\right)$
$K.E.=\frac{1}{2}mv^{2}$
$v^{2}=\frac{2KE}{m}$
$v=\sqrt{\frac{2KE}{m}}$
Substituting this in equation $\left(i\right)$
$\lambda =\frac{h}{mv} \sqrt{\frac{m}{2m\left(K.E\right)}}\,...\left(i\right)$
i.e. $\lambda\,\propto \frac{1}{\sqrt{KE}}$
$\therefore $ when $KE$ become 4 times wavelength become $1/2.$