Given, λ1=4972A˚ and λ2=6216A˚ and I=3.6×10−3Wm−2 Intensity associated with each wavelength =23.6×10−3=1.8×10−3Wm−2 Work function ϕ=hv=λhc =λ(6.62×10−34)(3×108)=λ12.4×103ev for different wavelengths ϕ1=λ112.4×103=497212.4×103=2.493eV=3.984×10−19 Jϕ2=λ212.4×103=621612.4×103=1.994eV=3.184×10−19 J Work function for metallic surface ϕ=2.3 eV (given) ϕ2<ϕ Therefore, ϕ2 will not contribute in this process. Now, no. of electrons per m2−s= no. of photons per m2−s no. of electrons per m2−s =3.984×10−191.8×10−3×10−4(∵1 cm2=10−4 m2)=0.45×1012 So, the number of photo electrons liberated in 2 sec. =0.45×1012×2 =9×1011
A beam of light has two wavelengths of 4972 Å and 6216 Å with a total intensity of 3.6 × 10^-3 Wm^-2 equally distributed among the two wavelengths.…
JEE Main 2014 — Physics Modern Physics
2014mcqhard
A beam of light has two wavelengths of 4972A˚ and 6216A˚ with a total intensity of 3.6×10−3 Wm−2 equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm2 of a clean metallic surface of work function 2.3eV. 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 s is approximately:
Official previous-year question
Held on 12 Apr 2014 · Verified 6 Jul 2026.
Options
- A
6×1011
- B
9×1011
- C
11×1011
- D
15×1011
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