Question

The de-Broglie wavelength of a neutron at ${972}^{o}C$ is $\lambda$.
What will be its wavelength at ${27}^{o}C$ ?

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

Answer 1

Answer: (C)

$E=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$

De - Broglie wavelength

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

Now for a molecule
$E=KT$ (Boltzmann Law)

$\lambda =\frac{h}{\sqrt{2mkT}}$

$\lambda =\frac{C}{\sqrt{T}}$

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_2}{T_1}}$

Here T is in K

$\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{300}{1245}}$

${\lambda }_2=2{\lambda }_1$