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{3 R T}{M}}=\sqrt{\frac{3 \times 8.314 \times 298}{4 \times 10^{-3}}}=1363 ms ^{-1} $
$ \lambda =\frac{h}{m v}=\frac{6.626 \times 10^{-34} \times 6.023 \times 10^{23}}{4 \times 10^{-3} \times 1363} $
$=7.3 \times 10^{-11}\, m$