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(B) Given the resistance formula $R = \frac{m_e v}{n e^2 L^2}$.From the de-Broglie wavelength relation,$\lambda = \frac{h}{p} = \frac{h}{m_e v}$.Given

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$10^4$

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(B) Given the resistance formula $R = \frac{m_e v}{n e^2 L^2}$.From the de-Broglie wavelength relation,$\lambda = \frac{h}{p} = \frac{h}{m_e v}$.Given $\lambda = L$,we have $L = \frac{h}{m_e v}$,which implies $m_e v = \frac{h}{L}$.Substituting this into the resistance formula: $R = \frac{(h/L)}{n e^2 L^2} = \frac{h}{n e^2 L^3}$.Since $n = \frac{N}{V} = \frac{N}{L^3}$ (where $N$ is the number of electrons),we get $R = \frac{h}{N e^2}$.For maximum resistance,the number of electrons $N$ must be minimum. The minimum number of electrons in a metallic sample is $N = 1$.Thus,$R_{max} = \frac{h}{e^2} = \frac{6.63 	imes 10^{-34}}{(1.60 	imes 10^{-19})^2}$.$R_{max} = \frac{6.63 	imes 10^{-34}}{2.56 	imes 10^{-38}} \approx 2.59 	imes 10^4 \, \Omega$.This value is closest to $10^4 \, \Omega$.

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