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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.
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)
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
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