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In a Photoelectric Experiment, the Collector Plate is at 2.0 V with Respect to the Emitter Plate Made of Copper (φ = 4.5 Ev). the Emitter is Illuminated by a Source

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प्रश्न

In a photoelectric experiment, the collector plate is at 2.0 V with respect to the emitter plate made of copper (φ = 4.5 eV). The emitter is illuminated by a source of monochromatic light of wavelength 200 nm. Find the minimum and maximum kinetic energy of the photoelectrons reaching the collector.

योग
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उत्तर

Given:-

Work function of copper, ϕ = 4.5 eV,

Wavelength of monochromatic light, λ = 200 nm

From Einstein's photoelectric equation, kinetic energy,

`K = E - phi = (hc)/lambda - phi`,

where, h = Planck's constant

c = speed of light

`therefore K = 1242/200 - 4.5`

`= 6.21 - 4.5 = 1.71  "eV"`

Thus, at least 1.7 eV is required to stop the electron. Therefore, minimum kinetic energy will be 2 eV.

It is given that electric potential of 2 V is required to accelerate the electron. Therefore, maximum kinetic energy

`= (2 + 1.7)  "eV" = 3.7  "eV"`

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अध्याय 42: Photoelectric Effect and Wave-Particle Duality - Exercises [पृष्ठ ३६६]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 42 Photoelectric Effect and Wave-Particle Duality
Exercises | Q 30 | पृष्ठ ३६६

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