$(a)\;3.68\times10^{-5}nm\qquad(b)\;7.38\times10^{-5}nm\qquad(c)\;7.38\times10^{-3}nm\qquad(d)\;6.35\times10^{-5}nm$

Mass of one helium atom = $\large\frac{4}{N_0}$

$=\large\frac{4}{6.02\times10^{23}}$

$V = \sqrt {\large\frac{3RT}{m}}$

$=\sqrt{\large\frac{3\times8.3143\times293}{4\times10^{-3}}}$m/sec

$\lambda = \large\frac{h}{mv} = \large\frac{6.626\times10^{-34}}{\large\frac{4}{6.02\times10^{23}}\times1351.69}$

$=7.38\times10^{-5}nm$

Hence the answer is (b)

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