Explanation
Given,
T=27+273=300K
{ \text { Boltzmann's constant. } }
{ k = 1.38 \times 10 ^ { - 23 } J m o l ^ { - 1 } k ^ { - 1 } }
Kinetic energy of neutron at absolute temperature
\text{ T is given by, }E=\dfrac{3}{2}kT
{ \lambda = \dfrac { h } { \sqrt { 2 m E } } }=\dfrac{h}{\sqrt{3mkT}}
{ \lambda = \dfrac { 6.63 \times 10 ^ { - 34 } } { \sqrt { 3 \times 1.675 \times 10 ^ { - 27 } \times 1.38 \times 10 ^ { - 23 } \times 300 } } }=1.45\times {{10}^{-10}}m\text{ }
Hence, De Broglie wavelength is 1.45\times {{10}^{-10}}m\text{ }
Given that,
Radius r={{r}_{0}}
We know that,
2\pi r=n\lambda
Suppose n=1
Then,
\lambda =2\pi {{r}_{0}}
Hence, this is the required solution
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