Question

The de-Broglie wavelength of a neutron at 27^oCisλ. What will be its wavelength at 927^oC

• A. λ / 2
• B. λ / 3
• C. λ/ 4
• D. λ / 9

Answer 1

Answer: (A)

λ / 2

Answer 2

$\lambda_{neutron} \propto \frac{1}{\sqrt{T}}$ ⇒ $\frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{T_{2}}{T_{1}}}$

⇒ $\frac{\lambda}{\lambda_{2}} = \sqrt{\frac{(273 + 927)}{(273 + 27)}} = \sqrt{\frac{1200}{300}} = 2$ ⇒ $\lambda_{2} = \frac{\lambda}{2}.$