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

The correct option is C $7.34\times {10}^{-11}m$

$V_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 298}{4\times {10}^{-3}}}=1363m{s}^{-1}$

$\lambda =\frac{h}{mv}=\frac{6.626\times {10}^{-34}\times 6.023\times {10}^{23}}{4\times {10}^{-3}\times 1363}$

$=7.32\times {10}^{-11}m$