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JAMB Physics 2008 Question 47

Question 47

Modern Physics

Caesium has a work function of 3 × 10−19J. The maximum energy of liberated electrons when it is illuminated by light of frequency 6.7 × 1014Hz is [h=6.6 × 10−34Js]

  1. A.1.42 × 10−19J.
  2. B.3.00 × 10−19J.
  3. C.4.42 × 10−19J.
  4. D.7.42 × 10−19J.
Show the answer and explanation

Answer: A (1.42 × 10−19J.)

Explanation

Photon energy hf = 6.6 × 10−34 × 6.7 × 1014 = 4.42 × 10−19 J. Maximum kinetic energy = hf − W = 4.42 × 10−19 − 3.00 × 10−19 = 1.42 × 10−19 J.