Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81v for the photoelectric emission experiment.
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Answer:Kmax = 1.3 × 10^-17 J and Vmax = 5.3 ×10^6 m/s
Explanation:
Kmax = eV° = 1.6 × 10^-19 × 81
= 1.29 × 10^-17 J
Vmax = 5.93 × 10^5 √V°
= 5.93 × 10^5 √81
= 5.3 × 10^6 m/s
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