Physics · Dual Nature Of Radiation And Matter · NEET
For the SAME energy E, a photon and an electron do NOT have the same momentum. A photon's momentum is p = E/c (light-like, so E = pc). An electron is a slow massive particle, so its momentum is p = root(2mE), which is far bigger. Memory hook: "Light is fast but light-weight in momentum; the electron is slow but momentum-heavy." So p(electron) is much greater than p(photon) at equal energy.
At equal energy, a photon uses p = E/c while an electron uses p = root(2mE); the electron's momentum is far larger, so it has the shorter de Broglie wavelength.
Your doubts, answered
For the same energy, which has more momentum, a photon or an electron?
The electron has much more momentum. Photon: p = E/c. Electron (non-relativistic): p = root(2mE). Divide them: p(electron)/p(photon) = c * root(2m/E). Since c is huge (3x10^8 m/s), this ratio is very large. Example (NEET 2026): for E = 20 eV, p(electron)/p(photon) = 225. So the electron wins by a big margin.
Why is the photon's momentum p = E/c but the electron's is p = root(2mE)?
They obey different energy-momentum rules. A photon has zero rest mass, so its full relation is E = pc, giving p = E/c. An electron has mass m and moves slowly, so its kinetic energy is E = p^2/(2m). Solve for p: p = root(2mE). Never use E = pc for an electron unless it is moving near the speed of light.
Does E = pc work for an electron too?
No, not at NEET speeds. E = pc is the light relation and holds only for a massless photon (or a particle moving so fast that rest energy is tiny). For a normal electron use E = p^2/(2m), which gives p = root(2mE). Mixing these up is the most common mistake in this topic.
How do I get the ratio of their momenta quickly?
Write both: p(photon) = E/c and p(electron) = root(2mE). Then p(electron)/p(photon) = root(2mE) / (E/c) = c*root(2mE)/E = c*root(2m/E). Plug numbers in SI units (E in joules, m in kg). This one line answers most exam questions.
How does momentum connect to de Broglie wavelength here?
Both a photon and an electron follow lambda = h/p. So a bigger momentum means a smaller wavelength. Because the electron has bigger momentum at the same energy, the electron has the SHORTER de Broglie wavelength, and the photon has the longer one. This is exactly what the 2016 and 2025 NEET questions test.
⚠️ The NEET trap ✗ For the same energy, the photon and electron have the same momentum because momentum = energy over speed for both. ✓ They differ. Photon: p = E/c (E = pc). Electron: p = root(2mE) from E = p^2/(2m). At equal energy the electron's momentum is much larger, by the factor c*root(2m/E). 🧠 See 'same energy' and reach for one formula? Stop. Photon uses E = pc; the massive electron uses E = p^2/(2m).
Real NEET questions
2026
A photon and an electron, each of 20 eV energy, move in free space. The ratio of the linear momentum of the electron p_e to that of the photon p_ph (p_e / p_ph) is: [c = 3x10^8 m/s, e = 1.6x10^-19 C, m_e = 9x10^-31 kg]
A · 2/450
B · 1/250
C · 225 ✓
D · 275
Solution: Step 1: Photon momentum. p_ph = E/c. Step 2: Electron momentum (non-relativistic). E = p^2/(2m), so p_e = root(2 m E). Step 3: Ratio. p_e/p_ph = root(2mE) / (E/c) = c * root(2m/E). Step 4: Put E = 20 eV = 20 x 1.6x10^-19 = 3.2x10^-18 J, m = 9x10^-31 kg. Inside root: 2m/E = (2 x 9x10^-31)/(3.2x10^-18) = 5.625x10^-13. root of that = 7.5x10^-7. Step 5: Multiply by c: 3x10^8 x 7.5x10^-7 = 225. Answer: 225 (option C).
2025
A photon and an electron (mass m) have the same energy E. The ratio lambda_photon / lambda_electron of their de Broglie wavelengths is: (c is the speed of light)
A · c*root(2m/E) ✓
B · (1/2c)*root(E/m)
C · root(2E/m)
D · 2c/root(mE)
Solution: Step 1: Photon wavelength. lambda_ph = h/p = hc/E (since p = E/c). Step 2: Electron wavelength. p = root(2mE), so lambda_e = h/root(2mE). Step 3: Take the ratio. lambda_ph/lambda_e = (hc/E) x (root(2mE)/h) = c*root(2mE)/E = c*root(2m/E). Answer: option A. Note this equals the momentum ratio p_e/p_ph, because lambda = h/p flips the ratio.
2016
An electron of mass m and a photon have the same energy E. The ratio of the de Broglie wavelengths associated with them (electron to photon) is (c being the velocity of light):
A · (1/c)*root(E/2m) ✓
B · root(E/2m)
C · c*root(2mE)
D · (1/c)*root(2m/E)
Solution: Step 1: Electron. lambda_e = h/root(2mE). Step 2: Photon. lambda_ph = hc/E. Step 3: Ratio electron to photon. lambda_e/lambda_ph = (h/root(2mE)) x (E/hc) = E/(c*root(2mE)) = (1/c)*root(E/2m). Answer: option A. This is just the reciprocal of the 2025 ratio, confirming lambda = h/p is consistent.
Solved Dual Nature Of Radiation And Matter NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
