Question

The de-Broglie wavelength of helium atom at room temperature is

• A. $6.6\times 10^{-34}m$
• B. $4.39\times 10^{-10}m$
• C. $7.34\times 10^{-11}m$
• D. $2.335\times 10^{-20}m$

Answer 1

Answer: (C)

$V_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 298}{4\times 10^{-3}}}=1363m{s}^{-1}$
$\lambda =\frac{h}{mv}=\frac{6.626\times 10^{-34}\times 6.023\times 10^{23}}{4\times 10^{-3}\times 1363}$
$=7.3\times 10^{-11}m$