Question

Consider a helium (He) atom that absorbs a photon of wavelength 330 nm. The change in the velocity (in cm s^-1) of the He atom after the photon absorption is_____.
(Assume: Momentum is conserved when the photon is absorbed.
Use: Planck constant = 6.6 × 10^-34 J s, Avogadro number = 6 × 1023 mol^-1, Molar mass of He = 4 g mol^-1)

Answer 1

Correct Answer: 30

Step 1: Calculate the energy of the absorbed photon:
The energy of the photon is given by the equation \( E = \frac{h c}{\lambda} \),
where:
\( h = 6.6 \times 10^{-34} \, \text{J s} \) (Planck's constant),
\( c = 3 \times 10^8 \, \text{m/s} \) (speed of light),
\( \lambda = 330 \, \text{nm} = 330 \times 10^{-9} \, \text{m} \) (wavelength of the photon).
Substitute the values:
\( E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{330 \times 10^{-9}} = 6.0 \times 10^{-19} \, \text{J} \).

Step 2: Calculate the momentum of the absorbed photon:
The momentum of a photon is given by the equation \( p = \frac{E}{c} \).
Substitute the values:
\( p = \frac{6.0 \times 10^{-19}}{3 \times 10^8} = 2.0 \times 10^{-27} \, \text{kg m/s} \).

Step 3: Use the conservation of momentum to find the change in velocity of the He atom:
The momentum change of the He atom is equal to the momentum of the photon.
The mass of a helium atom is \( m = \frac{4}{6 \times 10^{23}} \, \text{kg} = 6.67 \times 10^{-26} \, \text{kg} \).
Since momentum is conserved:
\( \Delta v = \frac{p}{m} = \frac{2.0 \times 10^{-27}}{6.67 \times 10^{-26}} = 3.0 \times 10^{-2} \, \text{m/s} = 30 \, \text{cm/s} \).

Step 4: Final Answer:
The change in the velocity of the He atom after the photon absorption is 30 cm/s.