Explanation

(c) de-Broglie wavelength of an electron is given by
$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}}$
Or $\lambda \propto \frac{1}{\sqrt{K}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $\frac{\lambda^{'}}{\lambda} = \frac{1}{\sqrt{3K}}\frac{\sqrt{K}}{1} = \frac{1}{\sqrt{3}}$
Or $\lambda^{'} = \frac{\lambda}{\sqrt{3}}$
Hence, de-Broglie wavelength will change by factor $\frac{1}{\sqrt{3}}$.