In a photoelectric experiment- the collector plate is at 2-0V with respect to the emitter plate made of copper - - 4-5 eV- the emitter is illuminated by a source of monochromatic light of wavelength 200 nm-

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Standard IX

Chemistry

Cathode Ray Experiment

In a photoele...

Question

In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper $(ϕ = 4.5 e V)$ . the emitter is illuminated by a source of monochromatic light of wavelength 200 nm.

The minimum kinetic energy of the photo electrons reaching the collector is 0

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The maximum kinetic energy of the photo electrons reaching collector is 3.7 eV.

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If the polarity of the battery is reversed then answer to part A will be 0

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If the polarity of the battery is reversed then answer to part B will be 1.7 eV

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Solution

The correct option is B The maximum kinetic energy of the photo electrons reaching collector is 3.7 eV.
$K = \frac{h c}{α} - ϕ$

$= \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{200 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - 4.5$

$= 1.7 e V$

Min. energy with which electrons emitted=1.7eV

Max. $K ε \Rightarrow K_{min} + e V_{0}$

$= 1.7 + 2$

$= 3.7 e V$

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In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper (ϕ=4.5eV). the emitter is illuminated by a source of monochromatic light of wavelength 200 nm.

The correct option is B The maximum kinetic energy of the photo electrons reaching collector is 3.7 eV.K=hcα−ϕ =6.63×10−34×3×108200×10−9×1.6×10−19−4.5 =1.7eV Min. energy with which electrons emitted=1.7eV Max. Kε⇒Kmin+eV0 =1.7+2 =3.7eV

In a photoelectric experiment, the collector plate is at 2.0V with respect to the emitter plate made of copper $(ϕ = 4.5 e V)$ . the emitter is illuminated by a source of monochromatic light of wavelength 200 nm.