Modern Physics 2 Question 37

36. A beam of light has three wavelengths 4144\AA,4972\AA and 6216\AA with a total intensity of 3.6×103Wm2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1.0cm2 of a clean metallic surface of work function 2.3eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds.

(1989, 8M)

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Solution:

  1. Energy of photon having wavelength 4144\AA,

E1=123754144eV=2.99eV

Similarly, E2=123754972eV=2.49eV and

E3=123756216eV=1.99eV

Since, only E1 and E2 are greater than the work function W=2.3eV, only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, intensity corresponding to each wavelength is

3.6×1033=1.2×103W/m2

Or energy incident per second in the given area ( A=1.0cm2=104m2) is

ρ=1.2×103×104=1.2×107J/s

Let n1 be the number of photons incident per unit time in the given area corresponding to first wavelength. Then

n1=ρE1=1.2×1072.99×1.6×1019=2.5×1011

Similarly, n2=ρE2=1.2×1072.49×1.6×1019

=3.0×1011

Since, each energetically capable photon ejects electron, total number of photoelectrons liberated in 2s.

=2(n1+n2)=2(2.5+3.0)×1011=1.1×1012



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