For a photon whose momentum is 10 MeV/c, where c is the speed of light, what is the energy of this photon expressed in MeV?

Introduction / Context:
In relativistic physics and quantum mechanics, photons are massless particles of light that carry energy and momentum. For photons, energy and momentum are related by a simple formula involving the speed of light. While massive particles obey the full relativistic energy momentum relation, photons satisfy E = p * c, where E is energy, p is momentum and c is the speed of light. This question gives the momentum of a photon in convenient units and asks you to find its energy in mega electron volts, which is a common energy unit in nuclear and particle physics.

Given Data / Assumptions:
• The particle is a photon, so its rest mass is zero. • Photon momentum p is given as 10 MeV/c. • c represents the speed of light in vacuum. • We are asked to express energy E in units of MeV.

Concept / Approach:
For photons, the energy momentum relation simplifies to E = p * c because the rest mass term in the general relativistic formula is zero. If momentum p is expressed in units of MeV/c, multiplying by c simply cancels c in the denominator, leaving energy in MeV. This is why high energy physics often quotes momentum in MeV/c or GeV/c, making conversion to energy straightforward for massless particles. Therefore, to find the photon energy, we substitute p = 10 MeV/c into E = p * c, giving E = (10 MeV/c) * c = 10 MeV.

Step-by-Step Solution:
Step 1: Write down the photon energy momentum relation: E = p * c. Step 2: Note that the given momentum is p = 10 MeV/c, already expressed in mixed units convenient for high energy physics. Step 3: Substitute p into the formula: E = (10 MeV/c) * c. Step 4: Recognise that the factor c in the numerator cancels the c in the denominator. Step 5: After cancellation, we obtain E = 10 MeV. Step 6: Conclude that the photon energy is 10 MeV, matching option A.

Verification / Alternative check:
If you prefer to think in SI units, you could convert 10 MeV to joules and compute p in kilogram metre per second, but this is unnecessary because the question uses natural high energy units. The key point is that MeV/c multiplied by c yields MeV. In many particle physics problems, energy and momentum are treated in these units specifically to simplify such calculations. Additionally, the general relativistic relation E^2 = p^2 * c^2 + m^2 * c^4 reduces to E = p * c when m = 0 for photons, confirming the starting formula. Both the short and long methods lead to the same result of 10 MeV.

Why Other Options Are Wrong:
Option B, 100 MeV, corresponds to mistakenly multiplying by an extra factor of ten, perhaps by misreading the given units or squaring c unnecessarily. Option C, 1 MeV, would result from dividing by ten instead of simply cancelling c. Option D, 0.1 MeV, represents an even larger error. None of these alternatives correctly apply the simple relation E = p * c when p is given in MeV/c. The units themselves signal that multiplication by c only cancels the denominator and does not introduce any additional numerical factor.

Common Pitfalls:
A common mistake is to overcomplicate the problem by trying to use the full relativistic energy equation, including rest mass, without noticing that photons have zero rest mass. Another pitfall is misunderstanding the meaning of MeV/c units and treating c as a simple number like 3 × 10^8 without respecting unit cancellation. Remember that in high energy physics, using MeV, MeV/c and MeV/c^2 as units is designed to simplify calculations. When you see p in MeV/c for a photon, multiplying by c immediately gives you E in MeV with no extra numerical factor.

Final Answer:
The correct choice is 10 MeV, because for a photon with momentum 10 MeV/c, the energy is E = p * c = (10 MeV/c) * c = 10 MeV.