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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.For a neutron in thermal equilibrium at temperature $T$,the kinetic energy is $E = \

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$\frac{30.7}{\sqrt{T}} \ \mathring{A}$

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(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.For a neutron in thermal equilibrium at temperature $T$,the kinetic energy is $E = \frac{p^2}{2m_n} = \frac{3}{2}kT$. However,using the approximation $E \approx kT$ as per the question:$p = \sqrt{2m_n E} = \sqrt{2m_n kT}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ J\cdot s$,$m_n = 1.675 	imes 10^{-27} \ kg$,and $k = 1.38 	imes 10^{-23} \ J/K$.$\lambda = \frac{h}{\sqrt{2m_n kT}} = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 1.675 	imes 10^{-27} 	imes 1.38 	imes 10^{-23} 	imes T}}$.$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{4.623 	imes 10^{-50} 	imes T}} = \frac{6.63 	imes 10^{-34}}{2.15 	imes 10^{-25} \sqrt{T}}$.$\lambda \approx \frac{3.08 	imes 10^{-9}}{\sqrt{T}} \ m = \frac{30.8}{\sqrt{T}} \ \mathring{A}$.Rounding to the nearest standard value,we get $\lambda = \frac{30.7}{\sqrt{T}} \ \mathring{A}$.

The de Broglie wavelength associated with a neutron at temperature $T$ is given by: $(E = kT)$

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