Question

The wavelength of neutron with a translatory kinetic energy equal to $kT$ at $400K$ is

• A. 153 pm
• B. 178 pm
• C. 17.8 pm
• D. 20.0 pm

Answer 1

Answer: (A)

Kinetic energy $=kT$

where, $k=$ Boltzmann's constant $=\frac{R}{N_0}$

Kinetic energy at $300K=\frac{RT}{N_0}$

$=\frac{8.314Jmo{l}^{-1}{K}^{-1}\times 400K}{6.02\times 10^{23}mo{l}^{-1}}$

$=5.524\times 10^{-21}J$

By de-Broglie equation, $\lambda =\frac{h}{\sqrt{2mKE}}$

$m$ (neutron) $=1.675\times 10^{-27}kg$

$\lambda =\frac{6.62\times 10^{-34}Js}{\sqrt{2\times 1.675\times 10^{-27}kg\times 5.524\times 10^{-21}J}}$

$=1.53\times 10^{-10}m=153\times 10^{-12}m$

$=153pm$

Note: $\left(1J=1kg{m}^2{s}^{-2}\right)$

$(\Delta x\cdot \Delta p{)}_{\text{electron }}\ge \frac{h}{4\pi }$

$(\Delta x{)}_{\text{helium }}=(\Delta x{)}_{\text{electron }}=\pm 1.00mm=\pm 1.00\times 10^{-3}m$

Hence, $(\Delta p{)}_{\text{helium }}=(\Delta p{)}_{\text{electron }}=5.0\times 10^{-26}kgm{s}^{-1}$