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I Multiple Choice Questions · Q4

Q.When a metallic surface is illuminated with radiation of wavelength λ\lambda, the stopping potential is VV. If the same surface is illuminated with radiation of wavelength 2λ2\lambda, the stopping potential is V4\dfrac{V}{4}. The threshold wavelength for the metallic surface is (NEET 2016):

(a) 4λ4\lambda
(b) 5λ5\lambda
(c) 52λ\dfrac{5}{2}\lambda
(d) 3λ3\lambda
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Step 1. Einstein's equation for wavelength λ\lambda gives eV=hcλ−ϕ0eV=\dfrac{hc}{\lambda}-\phi_0, and for wavelength 2λ2\lambda gives eV4=hc2λ−ϕ0\dfrac{eV}{4}=\dfrac{hc}{2\lambda}-\phi_0.

Step 2. Multiply the second equation by 4: eV=2hcλ−4ϕ0eV=\dfrac{2hc}{\lambda}-4\phi_0.

Step 3. Set this equal to the first equation's right side: hcλ−ϕ0=2hcλ−4ϕ0⇒3ϕ0=hcλ⇒ϕ0=hc3λ\dfrac{hc}{\lambda}-\phi_0=\dfrac{2hc}{\lambda}-4\phi_0 \Rightarrow 3\phi_0=\dfrac{hc}{\lambda} \Rightarrow \phi_0=\dfrac{hc}{3\lambda}. …

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