A) 0.565
B) 0.435
C) 0.223
D) 0.195
Correct Answer: A
Solution :
For a gaseous atom, a measure of its average speed is the root mean square (rms) speed as \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\]\[=\sqrt{\frac{3\times 8.314\times 500}{4\times {{10}^{-3}}}}=1766\,\text{m}{{\text{s}}^{-1}}\] Mass of a He-atom\[=\frac{4\times {{10}^{-3}}}{6.023\times {{10}^{23}}}=6.64\times {{10}^{-27}}kg\] \[\therefore \]de-Broglie wavelength \[(\lambda )\] \[=\frac{h}{mv}=\frac{6.626\times {{10}^{-34}}}{6.64\times {{10}^{-27}}\times 1766}\] \[=5.65\times {{10}^{-11}}m=0.565{\AA}\]
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