JEE Challenger
More from Structure of Atom
ChemMedAdv 2021 • Paper 2

Velocity Change of Helium Atom on Photon Absorption

Consider a helium (He\text{He}) atom that absorbs a photon of wavelength 330 nm330\text{ nm}. The change in the velocity (in cm s−1\text{cm s}^{-1}) of the He\text{He} atom following the photon absorption is ______.

(Assume: Momentum is conserved when the photon is absorbed. Use: Planck constant =6.6×10−34 J s= 6.6 \times 10^{-34}\text{ J s}, Avogadro number =6×1023 mol−1= 6 \times 10^{23}\text{ mol}^{-1}, Molar mass of He=4 g mol−1\text{He} = 4\text{ g mol}^{-1})

Official Numerical Answer30

Topics & Concepts

Structure of AtomBohr ModelDe Broglie Wavelength

Step-by-Step Solution

To find the change in velocity of the helium (He\text{He}) atom upon absorbing a photon, we apply the law of conservation of linear momentum.

  1. Calculate the momentum of the absorbed photon: The momentum of a photon (pphotonp_{\text{photon}}) is given by the de Broglie relation: pphoton=hλp_{\text{photon}} = \frac{h}{\lambda}

Given:

  • Planck's constant, h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s}
  • Wavelength of the photon, λ=330 nm=330×10−9 m=3.3×10−7 m\lambda = 330\text{ nm} = 330 \times 10^{-9}\text{ m} = 3.3 \times 10^{-7}\text{ m}

Substituting the values: pphoton=6.6×10−34 J s3.3×10−7 m=2.0×10−27 kg m s−1p_{\text{photon}} = \frac{6.6 \times 10^{-34}\text{ J s}}{3.3 \times 10^{-7}\text{ m}} = 2.0 \times 10^{-27}\text{ kg m s}^{-1}

  1. Calculate the mass of a single He\text{He} atom: The mass of one helium atom (mm) is: m=Molar mass of HeNAm = \frac{\text{Molar mass of He}}{N_A}

Given:

  • Molar mass of He=4 g mol−1=4×10−3 kg mol−1\text{He} = 4\text{ g mol}^{-1} = 4 \times 10^{-3}\text{ kg mol}^{-1}
  • Avogadro number, NA=6×1023 mol−1N_A = 6 \times 10^{23}\text{ mol}^{-1}

m=4×10−3 kg mol−16×1023 mol−1=23×10−26 kgm = \frac{4 \times 10^{-3}\text{ kg mol}^{-1}}{6 \times 10^{23}\text{ mol}^{-1}} = \frac{2}{3} \times 10^{-26}\text{ kg}

  1. Calculate the change in velocity (Δv\Delta v) of the He\text{He} atom: By conservation of linear momentum: ΔpHe=pphoton\Delta p_{\text{He}} = p_{\text{photon}} mΔv=hλm \Delta v = \frac{h}{\lambda} Δv=pphotonm\Delta v = \frac{p_{\text{photon}}}{m}

Substituting the values: Δv=2.0×10−27 kg m s−123×10−26 kg=3×10−1 m s−1=0.3 m s−1\Delta v = \frac{2.0 \times 10^{-27}\text{ kg m s}^{-1}}{\frac{2}{3} \times 10^{-26}\text{ kg}} = 3 \times 10^{-1}\text{ m s}^{-1} = 0.3\text{ m s}^{-1}

  1. Convert the change in velocity to cm s−1\text{cm s}^{-1}: Δv=0.3 m s−1×100 cm m−1=30 cm s−1\Delta v = 0.3\text{ m s}^{-1} \times 100\text{ cm m}^{-1} = 30\text{ cm s}^{-1}

Thus, the change in the velocity of the He\text{He} atom is 30.

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