NEET Physics Photoelectric Effect MCQ Question
If the work function of a metal is 3 eV and light of frequency 8 x 10^14 Hz is incident on it, what is the maximum kinetic energy of the emitted electrons? (Planck's constant, h = 6.63 x 10^-34 Js, 1 eV = 1.6 x 10^-19 J)
44 eV
44 eV
44 eV
0.44 eV
Correct Answer
Detailed Explanation
The maximum kinetic energy is calculated using K.E. = hν - φ where hν is the energy of the incident photon and φ is the work function. Substituting the values gives K.E. = (6.63 x 10^-34 * 8 x 10^14 / 1.6 x 10^-19) - 3 eV ≈ 2.44 eV.
Found an issue with this question?
Related Questions
More from Photoelectric Effect
More from
Match Column-I with Column-II. Column-I Column-II (a) de Broglie wavelength of an electron (i) 6.63 × 10^-34 J s (b) de Broglie wavelength...
What is the de Broglie wavelength of a particle with mass 0.15 kg moving at a speed of 30.0 m/s?
Which of the following statements about de Broglie wavelength is correct?