Studyledge

Quantum Physics: Question 5

Syllabus 22.1

Multiple choice A2 1 mark

A photon has energy E=3.00 eVE=3.00\text{ eV}.

What is the magnitude of the momentum of this photon?

Choose an answer to check it, then compare with the worked solution below.
Show worked solution Hide worked solution

Worked solution

Step 1: Convert the photon energy to joules

E=3.00×1.60×10−19=4.80×10−19 JE = 3.00\times1.60\times10^{-19} = 4.80\times10^{-19}\text{ J}

Step 2: Apply the photon momentum relation

A photon of energy EE has momentum: p=Ecp = \frac{E}{c}

Substituting E=4.80×10−19 JE=4.80\times10^{-19}\text{ J} and c=3.00×108 m s−1c=3.00\times10^{8}\text{ m s}^{-1}: p=4.80×10−193.00×108=1.60×10−27 kg m s−1p = \frac{4.80\times10^{-19}}{3.00\times10^{8}} = 1.60\times10^{-27}\text{ kg m s}^{-1}

Recompute as a check: 1.60×10−27×3.00×108=4.80×10−19 J1.60\times10^{-27}\times3.00\times10^{8}=4.80\times10^{-19}\text{ J}, which matches the photon energy in joules found in Step 1.

Why the other options are wrong

  • B (4.80×10−19 kg m s−14.80\times10^{-19}\text{ kg m s}^{-1}): this is just the photon energy in joules, taken as the momentum without dividing by cc at all.
  • C (1.44×10−10 kg m s−11.44\times10^{-10}\text{ kg m s}^{-1}): comes from multiplying by cc instead of dividing, i.e. computing EcEc rather than E/cE/c.
  • D (5.33×10−36 kg m s−15.33\times10^{-36}\text{ kg m s}^{-1}): comes from dividing by c2c^2 instead of cc, confusing the photon momentum relation with the mass-energy equivalence relation m=E/c2m=E/c^2.

Final answer

  • The photon’s momentum is 1.60×10−27 kg m s−1\boxed{1.60\times10^{-27}}\text{ kg m s}^{-1}, option A.