NEET 2016 Phase 2 · PhysicsEinstein eq.Previous Year Question
Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A if the stopping potential of A relative to C is:
Answer: (D) −3 V. Answer: (d) −3 V Solution: From first case: Kmax = E − φ ⇒ 2 = 5 − φ ⇒ φ = 3 eV.
- A.+3 V
- B.+4 V
- C.−1 V
- D.−3 V✓
Correct Answer
(D) −3 V
Solution & Explanation
Answer: (d) −3 V Solution: From first case: Kmax = E − φ ⇒ 2 = 5 − φ ⇒ φ = 3 eV. For 6 eV photons: Kmax = 6 − 3 = 3 eV. Stopping requires eV₀ = 3 eV, with A negative relative to C ⇒ stopping potential = −3 V.
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