Wavelength of photon absorbed, $$\lambda$$ = 330 nm = 330 $$\times$$ 10^$$-$$9 m

Planck's constant, h = 6.6 $$\times$$ 10^$$-$$34 J s

Molar mass of He, M = 4 g mol^$$-$$1 = 4 $$\times$$ 10^$$-$$3 kg mol^$$-$$1

Avogadro number, N_A = 6 $$\times$$ 10²³ mol^$$-$$1

Mass of one atom of He, $$m = {M \over {{N_A}}}$$

$$ = {{4 \times {{10}^{ - 3}}} \over {6 \times {{10}^{23}}}} = {2 \over 3} \times {10^{ - 26}}$$ kg

Velocity, = V cm/s.

Using de-Broglie equation,

$$\lambda = {h \over {mv}}$$

$$ \Rightarrow v = {h \over {m\lambda }} = {{6.6 \times {{10}^{ - 34}}} \over {2/3 \times {{10}^{ - 26}} \times 330 \times {{10}^{ - 9}}}}$$

$$ = {{6.6 \times 3 \times {{10}^{ - 34}} \times {{10}^{35}}} \over {2 \times 330}} = 0.03 \times 10 = 0.3$$ m/s

= 30 cm/s