Question

de-Broglie wavelength of a neutron at ${927}^{\circ }C$ is $\lambda$ . What will be its wavelength at ${27}^{\circ }C$?

• A. $\frac{\lambda }{9}$
  
• B. $\lambda$
  
• C. $2\lambda$
  
• D. $4\lambda$

Answer 1

The correct option is C $2\lambda$


de Broglie wavelength
Solution $\lambda =\frac{h}{\sqrt{3mKT}}$
$T_1=927+273=1200K{T}_2=273+27=300K{\lambda }_1=\frac{h}{\sqrt{3mK\times 1200}}{\lambda }_2=\frac{h}{3mK\times 300}\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{2}$