The beam of light has three wavelengths 4144A∘,4972A∘ and 6216A∘ with a total intensity of 3.6×10−3Wm2 equally distributed amongst the three wavelengths. The beam falls normally on the area 1cm2 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 2s.
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Correct option is B)
As we know, threshold wavelength (λ0)=ϕhc ⇒λ0=2.3×(1.6×10−19)(6.63×10−34)×3×108=5.404×10−7m. ⇒λ0=5404A∘ Hence, wavelength 4144A∘ and 4972A∘ will emit electron from the metal surface. For each wavelength energy incident on the surface per unit time = intensity of each × area of the surface wavelength =33.6×10−3×(1cm)2=1.2×10−7joule Therefore, energy incident on the surface for each wavelengths is 2s E=(1.2×10−7)×2=2.4×10−7J Number of photons n1 due to wavelength 4144A∘ n1=(6.63×10−34)(3×108)(2.4×10−7)(4144×10−10)=0.5×1012 Number of photon n2 due to the wavelength 4972A∘ n2=(6.63×10−34)(3×108)(2.4×10−7)(4972×10−10)=0.572×1012 Therefore total number of photoelectrons liberated in 2s, N=n1+n2 =0.5×1012+0.575×1012 =1.075×1012.
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