When a beam of 106 eV photon of intensity 2.0 W/m

Question

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 · Accepted Answer

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
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Modern Physics> When a beam of 106 eV photon of intensity...

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

Amit Saxena , 11 Years ago

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Modern Physics> When a beam of 106 eV photon of intensity...

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

Amit Saxena , 11 Years ago

Navjyot Kalra

LIVE ONLINE CLASSES

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