$(a)\;7.29\times10^{-11}m\qquad(b)\;4.28\times10^{-10}m\qquad(c)\;5.31\times10^{-11}m\qquad(d)\;6.28\times10^{-11}m$

$\lambda = \large\frac{h}{mu}$----(1)

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

$=\sqrt{\large\frac{3\times8.314\times300}{4\times10^{-3}}}$

=1367.8 m/Sec

Now substitute the value of u in(1)

$=\large\frac{6.626\times10^{-34}}{\large\frac{4\times10^{-3}\times1367.8}{6.023\times10^{23}}}$

$=7.29\times10^{-11}m$

Hence answer is (a)

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