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Calculate the wavelength of a helium atom whose speed is equal to root mean square speed at 293 K $(rms = \sqrt{3RT/m})$

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

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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)
answered Feb 4, 2014 by sharmaaparna1
 

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