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Two separate monochromatic light beams A and B of the same intensity (energy per unit ara per unit time) are falling normally on a unit area of a metallic respectively.Their wavelength are `lambda_(A)` and `lambda_(B)` respectively. Assuming that all the incident light is used in ejecting the photoelectrons form beam A and that from B is

A

`(lamda_(A)//lamda_(B))^(2)`

B

`(lamda_(B)//lamda_(A))^(2)`

C

`(lamda_(A)//lamda_(B))`

D

`(lamda_(B)//lamda_(A))1`

Text Solution

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The correct Answer is:
C
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Two separate monochromatic light beams A and B of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength are lambda_(A) and lambda_(B) respectively. Assuming that all the the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from B is

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  • If K_(1) and K_(2) are maximum kinetic energies of photoelectrons emitted when light of wavelength lambda_(1) and lambda_(2) respectively are incident on a metallic surface. If lambda_(1)=3lambda_(2) then

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  • Two metallic plates A and B , each of area 5 xx 10m are placed parallel to each other at a separation of 1 cm . Plate B carries a positive charge of 33.7 pc . A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0, so that 10 photons fall on it per square meter per second. Assume that one photoelectron is emitted for every 10 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV . Electric field between the plates at the end of 10 seconds is

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