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A metal plate of area 1×10−4m2 is illuminated by a radiation of intensity16 mW/m2. The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [1 eV=1.6×10−19J]

(a)

1010 and 5 eV

(b)

1014 and 10 eV

(c)

1012 and 5 eV

(d)

1011 and 5 eV


Question ID - 50271 | SaraNextGen Answer

A metal plate of area 1×10−4m2 is illuminated by a radiation of intensity16 mW/m2. The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [1 eV=1.6×10−19J]

(a)

1010 and 5 eV

(b)

1014 and 10 eV

(c)

1012 and 5 eV

(d)

1011 and 5 eV

1 Answer - 5876 Votes

3537

Answer Key : (d) -

I=

16×10−3= =1012



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