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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