Question

The energy of a neutron in eV whose de-Broglie wavelength is $1\stackrel{\text{0}}{\text{A}}$

• A. $1.67\times {10}^{-27}eV$
• B. $8.13\times {10}^{-2}eV$
• C. $6.62\times {10}^{-22}eV$
• D. $3.23\times {10}^{-2}eV$

Answer 1

The correct option is B $8.13\times {10}^{-2}eV$

Neutron $m=1.67\times {10}^{-27}$ kg

$l=\frac{h}{p}$ $\Rightarrow P=\frac{h}{\lambda }$ $=\frac{66\times {10}^{-34}}{1\times {10}^{-10}}$ kg m/s

$\Rightarrow P=6.6\times {10}^{-24}$ kg m/s

energy $E=\frac{P^2}{2m}=\frac{6.6\times 6.6\times {10}^{-48}}{2\times 1.67\times {10}^{-27}}$ joule

$=\frac{(6.6{)}^2}{3.34}\times {10}^{-21}\times \frac{1}{1.6\times {10}^{-19}}$ ev

E = 0.8 ev