Question

The de-Broglie wavelength of a neutron at ${927}^0$C is$\lambda$. Its wavelength at ${27}^0$C is:

• A. $\frac{\lambda }{2}$
• B. $\lambda$
• C. $2\lambda$
• D. $4\lambda$

Answer 1

The correct option is C $2\lambda$

$\lambda =\frac{h}{\sqrt{2mKT}}$
$T_1=927+273$ , $T_2=27+273$
$=1200K$ , $=300K$
$\lambda \alpha \frac{1}{\sqrt{T}}$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{T_2}}{\sqrt{T_1}}$
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\sqrt{300}}{\sqrt{1200}}$
${\lambda }_2=2\lambda$