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Q.

A metal plate of area 1×10-4m2 is illuminated by a radiation of intensity 16mWm-2. The work function of the metal is 5eV. The energy of the incident photons is 10eV and only 10% of it produces electrons. The number of emitted photoelectrons per second and their maximum energy, respectively will be 1eV=1.6×10-19J

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a

1010 and 5eV

b

1014 and 10eV

c

1012 and 5eV

d

1011 and 5eV

answer is C.

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Detailed Solution

If n photons, each with energy E, are incident on area A per second, 

Intensity I=nEA      n=IAE=16×10-3×1×10-410×1.6×10-19=1012 s-1

Number of emitted electrons per second is 10% of n=10100×1012=1011 s-1

Maximum kinetic energy of emitted photoelectrons Kmax=E-ϕ0=10-5 eV=5 eV.


 

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