Explanation
Given,
T=27+273=300K
Boltzmann's constant.
k=1.38×10−23Jmol−1k−1
Kinetic energy of neutron at absolute temperature
T is given by, E=32kT
λ=h√2mE=h√3mkT
λ=6.63×10−34√3×1.675×10−27×1.38×10−23×300=1.45×10−10m
Hence, De Broglie wavelength is 1.45×10−10m
Given that,
Radius r=r0
We know that,
2πr=nλ
Suppose n=1
Then,
λ=2πr0
Hence, this is the required solution
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