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JEE Main 201412 AprilOnline
Exam
JEE Main
Year
2014
Date
12 April
Mode
Online
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2014 · 12 April · Online · Physics · Dual Nature of Matter

A beam of light has two wavelengths of 4972 Å and 6216 Å with a...

A beam of light has two wavelengths of 4972A˚4972 Å and 6216A˚6216 Å with a total intensity of 3.6×10−33.6 \times 10^{-3} Wm−2\mathrm{Wm}^{-2} equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm21 \mathrm{~cm}^2 of a clean metallic surface of work function 2.3eV2.3 \mathrm{eV}. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s2 \mathrm{~s} is approximately:

Options

  1. A6×10116 \times 10^{11}
  2. B9×10119 \times 10^{11}
  3. C11×101111 \times 10^{11}
  4. D15×101115 \times 10^{11}

Correct answer

B. 9×10119 \times 10^{11}

Step-by-step solution

Given, λ1=4972A˚and\lambda_1=4972 Å and\lambda_2=6216 Åandand\mathrm{I}=3.6 \times 10^{-3} \mathrm{Wm}^{-2}Intensityassociatedwitheachwavelength=Intensity associated with each wavelength =\frac{3.6 \times 10^{-3}}{2}=1.81.8\times10−310^{-3}\mathrm{Wm}^{-2}WorkfunctionWork function\phi=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}=(6.62×10−34)(3×108)λ=12.4×103λev\begin{aligned}&=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\lambda}\\ &=\frac{12.4 \times 10^3}{\lambda} \mathrm{ev} \end{aligned}fordifferentwavelengthsfor different wavelengthsϕ1=12.4×103λ1=12.4×1034972=2.493eV=3.984×10−19 Jϕ2=12.4×103λ2=12.4×1036216=1.994eV=3.184×10−19 J\begin{array}{r} \phi_1=\frac{12.4 \times 10^3}{\lambda_1}=\frac{12.4 \times 10^3}{4972}=2.493 \mathrm{eV}\\ =3.984 \times10^{-19} \mathrm{~J}\\ \phi_2=\frac{12.4 \times 10^3}{\lambda_2}=\frac{12.4 \times 10^3}{6216}=1.994 \mathrm{eV}\\ =3.184 \times10^{-19} \mathrm{~J} \end{array}WorkfunctionformetallicsurfaceWork function for metallic surface\phi=2.3\mathrm{eV}(given)(given)\phi_2 \lt \phiTherefore,Therefore,\phi_2willnotcontributeinthisprocess.Now,no.ofelectronsperwill not contribute in this process. Now, no. of electrons per\mathrm{m}^2-\mathrm{s}=no.ofphotonsperno. of photons per\mathrm{m}^2-\mathrm{s}no.ofelectronsperno. of electrons per\mathrm{m}^2-\mathrm{s}=1.8×10−33.984×10−19×10−4(∵1 cm2=10−4 m2)=0.45×1012\begin{aligned}&=\frac{1.8 \times 10^{-3}}{3.984 \times 10^{-19}} \times10^{-4} \\ &\left(\because1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\right)=0.45 \times10^{12} \end{aligned}So,thenumberofphotoelectronsliberatedin2sec.So, the number of photo electrons liberated in 2 sec.=0.45 \times 10^{12} \times 2=9 \times 10^{11}$
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