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# When a beam of 106 eV photon of intensity 2.0 W/m2 falls on a platinum surface of area 1.0x 10-4 m2 and work function 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

Navjyot Kalra
7 years ago
Hello Student,
No. of photons /sec
= Energy incident on platinum surface per second / Energy of on photon
No. of photon incident per second
= 2 x 10 x 10-4 / 10. 6 x 1.6 x 10-19 = 1. 18 x 1014
As 0.53% of incident photon can eject photoelectrons
∴ No. of photoelectrons ejected per second
= 1.18 x 1014 x 0.53 / 100 = 6.25 x 1011
Minimum energy = 0 eV,
Maximum energy = (10.6 – 5.6 ) eV = 5eV
Thanks
Navjot Kalra