Question

I needed a solution to this problem: At 200°C, the velocity of hydrogen molecule is 2.4X10^5 cm/sec. What is the approximate de-Broglie wavelength in this case?

Answer 1

The de Broglie wavelength of a particle is
\[\lambda = \frac{h}{p} \]
Using nice SI units, that means
\[\lambda = \frac{6.626\times10^{-34} J\cdot s}{2\cdot 1.67\times10^{-27} kg\cdot 2.4\times10^{3} m/s} \approx 8.27\times10^{-11} m = 82.7 \space pm\]
where 1 pm = 1 picometer = 10^-12 m.