When photon of the energy 3.8 eV falls on metallic surface of work function 2.8 eV, then the kinetic energy of emitted electrons are

1 eV

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6.6 eV

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0 to 1 eV

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2.8 eV

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Solution

The correct option is A 1 eV
Photoelectric law
$K E = h ν - W h ν = 3.8 e V W = 2.8 e V K E = 3.8 - 2.8 = 1 e V$

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When photon of the energy 3.8 eV falls on metallic surface of work function 2.8 eV, then the kinetic energy of emitted electrons are

The correct option is A 1 eVPhotoelectric lawKE=hν−Whν=3.8eVW=2.8eVKE=3.8−2.8=1eV

The correct option is A 1 eV
Photoelectric law
$K E = h ν - W h ν = 3.8 e V W = 2.8 e V K E = 3.8 - 2.8 = 1 e V$