Dual Nature of Matter

JEE Main level, numerical answer, medium

148 of 200

Question 148 of 200

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630Å6630Å is 0.42V0.42V. If the threshold frequency is x×1013sx\times {10}^{13}s, where xx is (nearest integer): (Given, speed light =3×108ms−1=3\times {10}^{8}m{s}^{-1}. Planck's constant =6.63×10−34Js=6.63\times {10}^{-34}Js)

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Answer: 35

We know that, KEmax=hf−ϕ=hcλ−ϕK{E}_{max}=hf-\phi =\dfrac{hc}{\lambda }-\phi
Or, hcλ−ϕ=eV0\dfrac{hc}{\lambda }-\phi =e{V}_{0}, where, ϕ=hνth\phi =h{\nu }_{th} and V0{V}_{0} is stopping potential.
So, threshold frequency is
νth=cλ−eV0h{\nu }_{th}=\dfrac{c}{\lambda }-\dfrac{e{V}_{0}}{h}
=3×10866330×10−10−1.6×10−19×0.426.63×10−34=\dfrac{3\times {10}^{8}}{66330\times {10}^{-10}}-\dfrac{1.6\times {10}^{-19}\times 0.42}{6.63\times {10}^{-34}}
=36.630×1015−1.6×0.426.63×1015=\dfrac{3}{6.630}\times {10}^{15}-\dfrac{1.6\times 0.42}{6.63}\times {10}^{15}
=1015(36.630−1.6×0.426.63)=0.4524−0.1013={10}^{15}\left(\dfrac{3}{6.630}-\dfrac{1.6\times 0.42}{6.63}\right)=0.4524-0.1013
νth=35.11×1013{\nu }_{th}=35.11\times {10}^{13}

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