In a photoelectric experiment, doubling the *intensity* of incident light (frequency unchanged) will
Adouble the KE of photoelectrons
Bhave no effect on either current or KE
Chalve the photocurrent
Ddouble the photocurrent, KE unchanged
Answer & Solution
Correct answer: D. double the photocurrent, KE unchanged
Intensity = photons per unit area per second. More photons → more ejected electrons → higher photocurrent. But each photon still carries the same energy $h\nu$, so individual electron KE is unchanged. This is the hallmark of the quantum (not wave) behaviour of light.
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