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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 cm 2 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.

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. 

Grade:11

2 Answers

Kevin Nash
askIITians Faculty 332 Points
9 years ago
Hello Student,
Please find the answer to your question
E1 = 12400 / 4144 = 2.99 eV, E­2 = 12400 / 4972 = 2.49 eV,
E3 = 12400 / 6216 = 1.99 eV.
⇒ 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
∴ n1 = 1.2 x 10-7 / 2.99 x 1.6 x 10-19 = 2.5 x 1011
n2 = 1.2 x 10-7 / 2.99 x 1.6 x 10-19 = 3 x 1011
Total number of photons = 2 (n1 + n2)
= 3.01 x 1011 + 2.51 x 1011 = 5.52 x 1011
∴ Total number of photoelectrons ejected in two seconds = 11 x 1011
Thanks
Kevin Nash
askIITians Faculty
Dilip Diwan
36 Points
7 years ago
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