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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.

9 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

9 years ago
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