Question

De Broglie wavelength of 0.05 eV thermal neutron is-

• A. 1.3 Å
• B. 2 Å
• C. 5.4 Å
• D. 8 Å

Answer 1

Answer: (A)

1.3 Å

Answer 2

l = $\frac{h}{p}$= $\frac{h}{\sqrt{2m_{0}K}}$

l = $\frac{h}{\sqrt{2 \times 1.66 \times 10^{–27} \times 0.05 \times 1.6 \times 10^{–19}}}$

l = 1.28 Å

l $\approx$1.3 Å