# De-broglie wavelength of a neutron at 927 degree Celsius is lambda. what will be its wavelength at 27-degree Celsius?

$\lambda _{neutron}\alpha \frac{1}{\sqrt{T}}$

⇒ $\frac{\lambda _{1}}{\lambda _{2}} = \sqrt{\frac{T_{2}}{T_{1}}}$

⇒ $\frac{\lambda _{1}}{\lambda _{2}} = \sqrt{\frac{273+927}{273+27}}$

⇒ $\frac{\lambda _{1}}{\lambda _{2}} = \sqrt{\frac{1200}{300}}$

⇒ $\frac{\lambda _{1}}{\lambda _{2}} = 2$

λ2 = λ/2