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

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

Structure of Atom

Solution:

Kinetic energy $= kT$

where, $k =$ Boltzmann's constant $=\frac{R}{N_{0}}$

Kinetic energy at $300 K =\frac{R T}{N_{0}}$

$=\frac{8.314 \,J \,mol ^{-1} K ^{-1} \times 400 K }{6.02 \times 10^{23} mol ^{-1}}$

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

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

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

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

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

$=153 \,pm$

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

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

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

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