Explanation

(c) Kinetic energy of a particle at temperature $TK$ is $E = \frac{3}{2}kT$. The de-Broglie wavelength associated with it is
$\lambda = \frac{h}{\sqrt{2mE}} = \frac{h}{\sqrt{2m \times \frac{3}{2}kT}}$
$ie,\ \lambda \propto \frac{1}{\sqrt{T}}$
$\therefore\ \frac{\lambda_{927}}{\lambda_{27}} = \sqrt{\frac{27 + 373}{927 + 273}}$
$= \sqrt{\frac{300}{1200}} = \frac{1}{2}$
or $\lambda_{927} = \frac{\lambda_{27}}{2} = \frac{\lambda}{2}$