When you irradiate a metal with light of wavelength 421 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.01 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts
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Ответ:
A) m = F/a = 91.7/9.81 = 9.35 kg
B) m = F/a = 59.2/9.81 = 6.03 kg
C) m = F/a = 33.4/9.81 = 3.40 kg
D) m = F/a = 9.65/9.81 = 0.984 kg
Explanation: