A beam of light has three wavelengths 4144 Å, 4972

Question

A beam of light has three wavelengths 4144 Å, 4972 Å and 6216 Å with a total intensity of 3.6 x 10-3 W m-2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects on electron. Calculate the number of photo electrons liberated in two seconds.

Kevin Nash · Accepted Answer

Hello Student,Please find the answer to your questionE1 = 12400 / 4144 = 2.99 eV, E&shy;2 = 12400 / 4972 = 2.49 eV,E3 = 12400 / 6216 = 1.99 eV.&rArr; Only first two wavelengths are capable of ejecting photoelectrons.Energy incident per second= 3.6 / 3 x 10-3 x 10-4 = 1.2 x 10-7 J /s&there4; n1 = 1.2 x 10-7 / 2.99 x 1.6 x 10-19 = 2.5 x 1011n2 = 1.2 x 10-7 / 2.99 x 1.6 x 10-19 = 3 x 1011Total number of photons = 2 (n1 + n2)= 3.01 x 1011 + 2.51 x 1011 = 5.52 x 1011&there4; Total number of photoelectrons ejected in two seconds = 11 x 1011ThanksKevin NashaskIITians Faculty

Modern Physics> A beam of light has three wavelengths 414...

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Radhika Batra , 10 Years ago

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