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Physics MCQ

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JEE Physics

Why Doubling Light Intensity Doesn't Double Electron Energy

Go beyond formulas. Understand why Einstein's photoelectric effect revolutionized physics - and how it appears in JEE.

Diagram of photoelectric effect experimental setup showing an evacuated glass tube with a clean metal photocathode (emitter plate) on the left and an anode (collector plate) on the right. Monochromatic light rays (shown as parallel arrows with wavelength lambda) are incident on the photocathode from the left. A variable DC voltage source V is connected between the plates with positive terminal to collector and negative to emitter. An ammeter A is connected in series to measure photocurrent. Ejected photoelectrons are shown as small arrows moving from cathode to anode. The circuit is complete with clearly labeled components.

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Experimental setup for studying the photoelectric effect

When electromagnetic radiation of sufficiently high frequency strikes a metal surface, electrons are emitted from the surface. This phenomenon is called the photoelectric effect.

Key equation: Maximum kinetic energy of photoelectron

$K_{\text{max}} = h\nu - \phi$

where $h$ is Planck's constant $(6.63 \times 10^{-34} \text{J}\cdot\text{s}$ , $\nu$ is the frequency of incident light, and $\phi$ is the work function of the metal.

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If the intensity of incident light is doubled while keeping the frequency constant (above threshold), what happens to the maximum kinetic energy of emitted photoelectrons?