1. Use conservation of momentum

When the helium atom absorbs a photon, the atom gains the photon's momentum.

Photon momentum is: $p = \frac{h}{\lambda}$

So the change in velocity of the He atom is: $\Delta v = \frac{p}{m} = \frac{h}{\lambda m}$

2. Find mass of one He atom

Given:

• Molar mass of He $= 4\,\text{g mol}^{-1}$
• Avogadro number $= 6 \times 10^{23}\,\text{mol}^{-1}$

Mass of one He atom: $m = \frac{4\,\text{g}}{6 \times 10^{23}} = \frac{4 \times 10^{-3}\,\text{kg}}{6 \times 10^{23}}$ $m = \frac{4}{6} \times 10^{-26} = \frac{2}{3} \times 10^{-26}\,\text{kg}$ $m \approx 6.67 \times 10^{-27}\,\text{kg}$

3. Convert wavelength into SI units

Given: $\lambda = 330\,\text{nm} = 330 \times 10^{-9}\,\text{m} = 3.30 \times 10^{-7}\,\text{m}$

4. Calculate photon momentum

$p = \frac{h}{\lambda} = \frac{6.6 \times 10^{-34}}{3.3 \times 10^{-7}}$ $p = 2.0 \times 10^{-27}\,\text{kg m s}^{-1}$

5. Calculate change in velocity

$\Delta v = \frac{p}{m} = \frac{2.0 \times 10^{-27}}{6.67 \times 10^{-27}}$ $\Delta v \approx 0.30\,\text{m s}^{-1}$

6. Convert to cm s$^{-1}$

$0.30\,\text{m s}^{-1} = 30\,\text{cm s}^{-1}$

Final Answer

$\boxed{30}$