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I Multiple Choice Questions · Q6

Q.A photoelectric surface is illuminated successively by monochromatic light of wavelength λ\lambda and λ2\dfrac{\lambda}{2}. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface material is (NEET 2015):

(a) hcλ\dfrac{hc}{\lambda}
(b) 2hcλ\dfrac{2hc}{\lambda}
(c) hc3λ\dfrac{hc}{3\lambda}
(d) hc2λ\dfrac{hc}{2\lambda}
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Step 1. At wavelength λ\lambda: K1=hcλ−ϕ0K_1=\dfrac{hc}{\lambda}-\phi_0. At wavelength λ/2\lambda/2 (double the frequency): K2=hcλ/2−ϕ0=2hcλ−ϕ0K_2=\dfrac{hc}{\lambda/2}-\phi_0=\dfrac{2hc}{\lambda}-\phi_0.

Step 2. Given K2=3K1K_2=3K_1: 2hcλ−ϕ0=3(hcλ−ϕ0)=3hcλ−3ϕ0\dfrac{2hc}{\lambda}-\phi_0=3\left(\dfrac{hc}{\lambda}-\phi_0\right)=\dfrac{3hc}{\lambda}-3\phi_0.

Step 3. Rearranging: 2ϕ0=3hcλ−2hcλ=hcλ2\phi_0=\dfrac{3hc}{\lambda}-\dfrac{2hc}{\lambda}=\dfrac{hc}{\lambda}, so ϕ0=hc2λ\phi_0=\dfrac{hc}{2\lambda}. …

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