Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s$$^{-1}$$) of He atom after the photon absorption is ___.
(Assume: Momentum is conserved when photon is absorbed.
Use: Planck constant = $$6.6 \times 10^{-34}$$ J s, Avogadro number = $$6 \times 10^{23} mol^{-1}$$, Molar mass of He = 4 g mol$$^{-1}$$)

The incident photon is completely absorbed by a stationary He atom, so linear momentum is conserved.
Initial momentum of the system = momentum of photon = $$\frac{h}{\lambda}$$ and final momentum of the system = momentum of the recoiling He atom = $$m_{\text{He}}\Delta v$$.

According to conservation of momentum (no external force),
$$m_{\text{He}}\Delta v = \frac{h}{\lambda}$$ $$-(1)$$

Photon momentum
$$h = 6.6 \times 10^{-34}\,\text{J s}$$
$$\lambda = 330\,\text{nm} = 330 \times 10^{-9}\,\text{m} = 3.3 \times 10^{-7}\,\text{m}$$
Therefore,
$$\frac{h}{\lambda} = \frac{6.6 \times 10^{-34}}{3.3 \times 10^{-7}} = 2.0 \times 10^{-27}\,\text{kg m s}^{-1}$$

Mass of one He atom
Molar mass $$= 4\,\text{g mol}^{-1} = 4 \times 10^{-3}\,\text{kg mol}^{-1}$$
Avogadro number $$N_A = 6 \times 10^{23}\,\text{mol}^{-1}$$
$$m_{\text{He}} = \frac{4 \times 10^{-3}}{6 \times 10^{23}} = 0.6667 \times 10^{-26}\,\text{kg} = 6.667 \times 10^{-27}\,\text{kg}$$

Change in velocity of He atom
From $$(1)$$,
$$\Delta v = \frac{h/\lambda}{m_{\text{He}}} = \frac{2.0 \times 10^{-27}}{6.667 \times 10^{-27}} = 0.30\,\text{m s}^{-1}$$

Convert to centimetres per second:
$$0.30\,\text{m s}^{-1} \times 100 = 30\,\text{cm s}^{-1}$$

Hence, the change in velocity of the He atom after absorbing the photon is 30 cm s$$^{-1}$$.