Question

If the mass of neutron $=1.7\times 10^{- 27}\text{ kg}$ , then the de-Broglie wavelength of neutron of energy $\text{3 eV}$ is $\left(\right.h=6.6\times \left(10\right)^{- 34 \, }J-s\left.\right)$

• A. $1.6\times 10^{- 16}m$
• B. $1.6\times 10^{- 11}m$
• C. $1.4\times 10^{- 10}m$
• D. $1.4\times 10^{- 11}m$

Answer 1

Answer: (B)

Given $E=3 \, eV=3\times 1.6\times 10^{- 19}$ J
We know that $\lambda =\frac{h}{\sqrt{2 m E}}$
$=\frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 1.7 \times 10^{- 27} \times 3 \times 1.6 \times 10^{- 19}}}=1.65\times 10^{- 11 \, }m$