Dual Nature of Matter

JEE Main level, single correct, medium

184 of 200

Question 184 of 200

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:

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Answer: (B) 9×10119 \times 10^{11}

Given, λ1=4972A˚\lambda_1=4972 Å
and λ2=6216A˚\lambda_2=6216 Å
and I=3.6×10−3Wm−2\mathrm{I}=3.6 \times 10^{-3} \mathrm{Wm}^{-2}
Intensity associated with each wavelength
=3.6×10−32=1.8×10−3Wm−2=\frac{3.6 \times 10^{-3}}{2}=1.8 \times 10^{-3} \mathrm{Wm}^{-2}
Work function ϕ=hv=hcλ\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}

for different wavelengths
ϕ1=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 \times 10^{-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 \times 10^{-19} \mathrm{~J} \end{array}
Work function for metallic surface ϕ=2.3\phi=2.3 eV\mathrm{eV} (given)
ϕ2<ϕ\phi_2 \lt \phi
Therefore, ϕ2\phi_2 will not contribute in this process.
Now, no. of electrons per m2−s=\mathrm{m}^2-\mathrm{s}= no. of photons per m2−s\mathrm{m}^2-\mathrm{s}
no. of electrons per m2−s\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}} \times 10^{-4} \\ &\left(\because 1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\right)=0.45 \times 10^{12} \end{aligned}
So, the number of photo electrons liberated
in 2 sec.
=0.45×1012×2=0.45 \times 10^{12} \times 2
=9×1011=9 \times 10^{11}

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