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Problems · Problem 2.14

Q.Calculate the mass of a photon with wavelength 3.6 Å.

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A photon has no rest mass, but carries relativistic mass m=hλcm = \frac{h}{\lambda c} due to its energy. For λ=3.6 A˚\lambda = 3.6\,\text{Å}, the relativistic mass is 6.13×10−32 kg\boxed{6.13 \times 10^{-32}\,\text{kg}}.

The Concept: Photon Mass and de Broglie's Relation

Photons are peculiar. They have zero rest mass—if you could somehow "catch" a photon and weigh it at rest, your scale would read zero. But photons are never at rest; they always travel at speed cc, and they carry energy. Einstein's mass-energy equivalence E=mc2E = mc^2 tells us that energy has an associated mass, called the relativistic mass or effective mass.

De Broglie's wavelength formula connects a particle's momentum to its wavelength:

λ=hp\lambda = \frac{h}{p}

where h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34}\,\text{J·s} is Planck's constant and pp is momentum. For a photon, momentum relates to energy by E=pcE = pc, and since E=mc2E = mc^2 for the relativistic mass, we have p=mcp = mc. Combining these gives us a direct route from wavelength to mass.


Step-by-Step Calculation

  1. Start with de Broglie's relation for momentum The momentum of any particle (including a photon) is:

p=hλp = \frac{h}{\lambda}

  1. Relate photon momentum to relativistic mass For a photon moving at speed cc, the momentum is also:

p=mcp = mc

where mm is the relativistic mass we seek.

  1. Equate the two expressions for momentum

mc=hλmc = \frac{h}{\lambda}

Solving for mm:

m=hλcm = \frac{h}{\lambda c}

m=hλcm = \frac{h}{\lambda c}

  1. Convert the wavelength to SI units

    Given λ=3.6 A˚=3.6×10−10 m\lambda = 3.6\,\text{Å} = 3.6 \times 10^{-10}\,\text{m}.

  2. Substitute the values

m=6.626×10−34(3.6×10−10)(3×108)m = \frac{6.626 \times 10^{-34}}{(3.6 \times 10^{-10})(3 \times 10^8)}

Calculate the denominator: …

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