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49.ONE milliw...

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49.ONE milliwatt of light of wavelength 4560 amstrong is incident on a cesium surface of work function 1.9 electronvolt.Given that quantum of efficiency of photoelectric emission is 0.5 percentage . The photoelectic current liberated as:

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Q. One milliwatt of light of wavelength

λ = 4560 o A

is incident on a cesium metal surface. Calculate the electron current liberated. Assume a quantum efficiency of

η = 0.5 %

. [Work function for cesium

= 1.89 e V

h c = 12400 e V - A

Q. One milliwatt of light of wavelength

4560 ˚ A

is incident on a caesium surface. Calculate the electron current liberated. Assume a quantum efficiency of

0.5 %

Q. One milliwatt of light of wavelength

λ = 4560 ˚ A

is incident on a cesium metal surface. Calculate the electrons liberaated. Assume a quantum efficiency of

η = 0.5 %

. [work function for cesium

= 1.89 e V

h c = 12400 e V - ˚ A

Q. Light of wavelength 5000

A^{0}

is incident on a metallic surface of work function 3.31 eV. Then photoelectric emission is

Q. The work function of a metal is 1eV. On making light of wavelength

3000 ˚ A

incident on this metal, the velocity of photoelectrons emitted from it for photoelectric emission will be

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