NEET 2016 Phase 2 · PhysicsPhoton energyPrevious Year Question
Electrons of mass m with de Broglie wavelength λ fall on the target in an X-ray tube. The cutoff wavelength (λ₀) of the emitted X-ray is:
Answer: (A) λ₀ = 2mcλ²/h. Answer: (a) λ₀ = 2mcλ²/h Solution: Electron momentum p = h/λ, so kinetic energy E = p²/2m = h²/2mλ².
- A.λ₀ = 2mcλ²/h✓
- B.λ₀ = 2h/mc
- C.λ₀ = 2m²c²λ³/h²
- D.λ₀ = λ
Correct Answer
(A) λ₀ = 2mcλ²/h
Solution & Explanation
Answer: (a) λ₀ = 2mcλ²/h Solution: Electron momentum p = h/λ, so kinetic energy E = p²/2m = h²/2mλ². At cutoff, all this energy becomes one photon: E = hc/λ₀. h²/2mλ² = hc/λ₀ ⇒ λ₀ = 2mcλ²/h.
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