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pooja paliwal Grade: 11
        












When a beam of 10.6 eV photons of intensity 2.0 W/m2 falls on a platinum surface of area 1.0 * 10 -4 m2 and work functions 5.6 eV. 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energies (in eV). Take 1 eV = 1.6 * 10 -19 J.

8 years ago

Answers : (1)

Gaurav Sharma
19 Points
										














Energy of incident photon,


 


                Ei = 10.6 eV


                   =10.6 × 1.6 × 10–19 J


                   = 16.96 × 10–19 J


 


        Energy incident per unit per unit time (intensity) = 2J


 


        \ Number of photons incident on unit area in unit time


 


                        = 2/16.96 * 10 -19


 


                        = 1.18 × 1014


 


Therefore, number of photons incident per unit time on given area

(1.0 × 10–4 m2)


 


                = (1.18 × 1018) (1.0 × 10–4)


 


                = 1.18 × 1014


 


But only 0.53% of incident photons emit photoelectrons


 


\ Number of photoelectrons emitted per second (n)


 


                n = (0.53/100) (1.18 × 1014)


 


                n = 6.25 × 1011


 


                Kmin = 0


 


and   Kmax = Ei – work function


 


                = (10.6 – 5.6) eV = 5.0 eV


 


\      Kmax = 5.0 eV

8 years ago
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