0

Guest

Courses New

When a beam of 106 eV photon of intensity 2.0 W/m 2 falls on a platinum surface of area 1.0x 10 -4 m 2 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

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

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Hello Student,
Please find the answer to your question
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
askIITians Faculty

Think You Can Provide A Better Answer ?

Other Related Questions on Modern Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More